Solutions – Upper Primary DivisionSolutions

Solutions – Upper Primary Division

1. (Also (Also MP2)MP2)

The numbers in order are 555, 556, 565, 566, 655, The numbers in order are 555, 556, 565, 566, 655,

hence (D). hence (D).

2. (Also (Also MP6)MP6)

One pizza will have 4 quarters, so two pizzas will have 2

One pizza will have 4 quarters, so two pizzas will have 2××4 = 8 quarters,4 = 8 quarters,

hence (D). hence (D). 3. 3. • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • P P QQ R R

The number of interior lattice points is 1 + 2 + 3 + 4 + 5 = 15, The number of interior lattice points is 1 + 2 + 3 + 4 + 5 = 15,

hence (C). hence (C). 4. 4. 00..3 + 03 + 0..4 4 = = 00..7,7, hence (B). hence (B). 5. 5. (Also (Also MP10)MP10) Alternative 1 Alternative 1 In cents, 500

In cents, 500÷÷80 = 6r20 so that he buys 6 chocolates and has 20 cents left,80 = 6r20 so that he buys 6 chocolates and has 20 cents left,

hence (C). hence (C).

Alternative 2 Alternative 2

Multiples of 80 are 80

Multiples of 80 are 80,,160160,,240240,,320320,,400400,,480480,,560. 560. FFrom thisrom this, he , he can aﬀorcan aﬀord 6 d 6 chchoco-oco- lates but not 7,

lates but not 7,

hence (C). hence (C). 6. 6. (Also (Also MP9)MP9) 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10

The opposite chair is both 5 places forward and 5 places back. The opposite chair is both 5 places forward and 5 places back. Five places back from chair 9 is chair 4,

Five places back from chair 9 is chair 4,

hence (D). hence (D).

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7. _IIn_{n oorrd}_deerr,_, _iis_{s w}_woorrtth_{h 4}_{4 +}_{+ 2}_{2 +}_{+ 1}_{1 +}₊ 22

2 2 = 8 beats, = 8 beats, hence (E). hence (E). 8. 8. _(Also_{(Also MP13)}_MP13)

She either has a 50c coin or not. She either has a 50c coin or not.

If she has a 50c coin, then she has one other 10c coin: 50 + 10 = 60. If she has a 50c coin, then she has one other 10c coin: 50 + 10 = 60. If she has no 50c coins, then she either has 0, 1, 2 or 3 20c coins: If she has no 50c coins, then she either has 0, 1, 2 or 3 20c coins:

220 0 + + 220 0 + + 220 0 = = 660 0 220 0 + + 220 0 + + 110 0 + + 110 0 = = 6600 220 0 + + 10 10 + + 10 10 + + 10 10 + + 10 10 = = 60 60 110 0 + + 110 0 + + 110 0 + + 110 0 + + 110 0 + + 110 0 = = 6600 In all, there are 5 possibilities,

In all, there are 5 possibilities,

hence (D). hence (D).

9. _(A)_{(A) holds}_holds 11

3 of of 30030000 mL whimL which is 10ch is 100000 mLmL..

(B) holds

(B) holds 33₄₄ of of 10010000 mL whimL which is 75ch is 7500 mLmL.. (C) holds

(D) holds 11₃₃ of 750 of 750 mL whmL whicich is 250h is 250 mLmL.. (E) holds

(E) holds 11₄₄ of of 20020000 mL whimL which is 50ch is 5000 mL.mL. Of these 1000

Of these 1000 mL is the mL is the greagreatesttest,,

hence (A). hence (A).

10.

10._{Here are the axes of symmetry of each shape:}_{Here are the axes of symmetry of each shape:}

no o aaxxees s 6 6 aaxxees s 8 8 aaxxees s nno o aaxxees s 113 3 aaxxeess

hence (E). hence (E).

11.

11._{(Also MP15)}_{(Also MP15)}

Starting from the outer end of the spiral (the loop on the rope) the dark and light Starting from the outer end of the spiral (the loop on the rope) the dark and light sections are longest, and the light sections are of similar length to the dark sections. sections are longest, and the light sections are of similar length to the dark sections. As you move towards the other end of the rope, both dark and light sections get As you move towards the other end of the rope, both dark and light sections get shorter. Only rope (A) shows this,

shorter. Only rope (A) shows this,

hence (A). hence (A).

2016 AMC – Upper Primary Solutions 2016 AMC – Upper Primary Solutions 38

12.

12._{(Also MP18)}_{(Also MP18)}

The larg

The large square me square must havust have side 8e side 8 cm. cm. Then alThen all of l of the cornthe corners of the pieces in theers of the pieces in the tan

tangragram lie on a m lie on a grigrid of d of 22 cmcm××22 cm squcm squaresares..

The shaded square has the same area as 2 of the grid squares, or 2

The shaded square has the same area as 2 of the grid squares, or 2××(2cm(2cm××22 cmcm) =) =

88 cmcm22_,, hence (D). hence (D). 13. 13. Alternative 1 Alternative 1 Since 8

Since 8××25 = 200, he needs to make 8 batches. This requires 825 = 200, he needs to make 8 batches. This requires 8××2211

2 2 = 20 packets = 20 packets of chocolate chips, of chocolate chips, hence (A). hence (A). Alternative 2 Alternative 2 Ea

Each pacch packeket t of of chchocolocolate chiate chips ps is enouis enough gh for 10 for 10 bibiscscuiuitsts. . So So for 200 for 200 bibiscscuiuits, ts, 2020 packets are required,

packets are required,

hence (A). hence (A).

14.

14._{Two even numbers add to even. For example, 2 + 4 = 6. So, not (A).}_{Two even numbers add to even. For example, 2 + 4 = 6. So, not (A).}

The diﬀerence betwee

The diﬀerence between two odds n two odds is alwais always even. ys even. FFor example, 5or example, 5−−1 = 4. So, not (B).1 = 4. So, not (B).

The sum of two odd numbers is always even. For example, 5 + 1 = 6. So, not (C). The sum of two odd numbers is always even. For example, 5 + 1 = 6. So, not (C). Ad

Addiding ng 3 3 odd numodd numbers is bers is the samthe same e as odd as odd plplus evenus even, , whwhicich h is odd. is odd. FFor examor examplple,e, 3 +

3 + 5 + 9 = 8 + 5 + 9 = 8 + 9 = 17. 9 = 17. So (D) So (D) is true.is true.

Two odd numbers multiplied is always odd. For example, 3

Two odd numbers multiplied is always odd. For example, 3××7 = 21. So, not (E),7 = 21. So, not (E),

hence (D). hence (D).

15.

15._{The side of the outer square is 36}_{The side of the outer square is 36}÷÷ 4 4 = = 99 cm and the sicm and the side of de of ththe e ininnener r sqsquauare isre is

20÷÷4 = 5cm.4 = 5cm.

The di

The diﬀereﬀerence is 4nce is 4 cm, whicm, which is 2ch is 2 cm on eaccm on each side. h side. So eacSo each rectanh rectangle is 7gle is 7 cmcm××2cm,2cm,

wit

with perimeteh perimeter 18r 18 cm,cm,

hence (E). hence (E).

16.

16. Alternative 1 Alternative 1

The possibilities are The possibilities are

Fiirrsst t nnuummbbeer r PPoossssiibbiilliittiiees s CCoouunntt

2 2 2 2113344, , 2211443 3 22 3 3 33112244, , 33114422, , 33221144, , 3322441 1 44 4 4 44112233, , 44113322, , 44221133, , 44223311, , 44331122, , 4433221 1 66 12 12 hence (B). hence (B).

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Alternative 2 Alternative 2

Ignoring whether the ﬁrst number is larger or smaller than the second, the number Ignoring whether the ﬁrst number is larger or smaller than the second, the number of combinations is 4

of combinations is 4××33××22 ××1 = 24.1 = 24.

Half of these will have the ﬁrst number larger than the second, and half will be the Half of these will have the ﬁrst number larger than the second, and half will be the other way around. So there are 12 combinations,

other way around. So there are 12 combinations,

hence (B). hence (B).

17.

17._{Each of (A), (B), (C) and (D) is possible:}_{Each of (A), (B), (C) and (D) is possible:}

However (E) is not possible. It has a total of 7 sides, whereas the hexagon has 6 sides However (E) is not possible. It has a total of 7 sides, whereas the hexagon has 6 sides and a cut increases the number of sides by at least 2, giving 8 or more sides,

and a cut increases the number of sides by at least 2, giving 8 or more sides,

hence (E). hence (E).

18.

18._{As the sum 3 + 9 + 15 + 18 + 24 + 29 = 98, we are looking for two groups of three}_{As the sum 3 + 9 + 15 + 18 + 24 + 29 = 98, we are looking for two groups of three}

numbermbers each with a s each with a sum clossum close to e to 49. 49. TherThere e are no are no threthree of e of the numthe numbers addinbers adding tog to 49, so the diﬀerence must be 2 or more.

49, so the diﬀerence must be 2 or more.

To get the diﬀerence as close as possible to 0, each sum will be as close to 49 as To get the diﬀerence as close as possible to 0, each sum will be as close to 49 as possible.

possible.

Noting that 3+18+29 = 50 and 9+15+24 = 48 gives (3+18+29)

Noting that 3+18+29 = 50 and 9+15+24 = 48 gives (3+18+29)−−(9+15+24) = 2,(9+15+24) = 2,

hence (C). hence (C).

19.

19._{Adding the upward-sloping diagonal, 4 + 7 + 10 + 13 = 34 is the common total.}_{Adding the upward-sloping diagonal, 4 + 7 + 10 + 13 = 34 is the common total.}

In th

In the 4th coe 4th colulumn 13mn 13 ++ 1212 ++ 1 = 26 an1 = 26 and 34d 34−−26 26 = 8. = 8. In tIn the 2he 2nd rnd roow, 5w, 5 ++ 1010 ++ 8 = 28 = 233

and 34

and 34−−23 = 11. Place23 = 11. Place AA andand BB in the third row as shown. in the third row as shown.

4 4 11 7 7 1212 5 5 1010 X X ₁₃₁₃ 8 8 11 11 A A BB

From the ﬁrst column,

From the ﬁrst column, AA₊₊ X X ₌_{= 34}₃₄₋₋_{9 = 25. From the downward-sloping diagonal,}_{9 = 25. From the downward-sloping diagonal,} B

B ₊₊X X ₌_{= 34}₃₄₋₋_{12 = 22. From the third row,}_{12 = 22. From the third row,} AA₊₊ BB ₌_{= 34}₃₄₋₋_{19 = 15.}_{19 = 15.}

Then 25 + 22 + 15 = 62 counts each of

Then 25 + 22 + 15 = 62 counts each of AA_,, BB _and_and X X _{twice, so that}_{twice, so that} AA₊₊ BB ₊₊X X ₌_{= 31}₃₁

and and X X _{= 31}₌₃₁₋₋_{15 = 16,}_{15 = 16,} hence (A). hence (A). 20. 20._{(Also J18)}_{(Also J18)} R R B B GG G G R R R R G G G G First try not to use a blue counter at all.

First try not to use a blue counter at all. The counters can’

The counters can’t be t be all red or all green, all red or all green, so start with a so start with a red counterred counter at the bottom of the

at the bottom of the circircle. cle. By the ﬁrst rule, the counBy the ﬁrst rule, the counters eitters eitherher side must be green.

side must be green.

Then, by the second rule, the counters opposite these green counters Then, by the second rule, the counters opposite these green counters must be red. The top counter can now be coloured green.

2016 AMC – Upper Primary Solutions 2016 AMC – Upper Primary Solutions 40

The side coun

The side counters cannoters cannot t be be red because they are adjacenred because they are adjacent t to a to a red. red. HoHowewevever, onlyr, only one of

one of them can be them can be greegreen, so n, so the other musthe other must t be be blublue. e. ThiThis arranges arrangemenment, as t, as shoshown,wn, satisﬁes the three rules.

satisﬁes the three rules.

Hence, the minimum number of blue counters is 1, Hence, the minimum number of blue counters is 1,

hence (B). hence (B).

21.

21._{(Also MP23)}_{(Also MP23)}

The discs are back in their original positions after 5 moves. The discs are back in their original positions after 5 moves.

rr b b gg y y oo 11 − −→→ gg y y oo rr b b 22 − −→→ oo rr b b gg y y 33 − −→→ b b gg y y oo rr 44 − −→→ y y oo rr b b gg 55 − −→→ rr b b gg y y oo

They will be in their original positions again after 10, 15 and 20 moves. After 1 more They will be in their original positions again after 10, 15 and 20 moves. After 1 more move, blue will be on the bottom,

move, blue will be on the bottom,

hence (B). hence (B). 22. 22._{(Also MP24)}_{(Also MP24)} Alternative 1 Alternative 1

Replacing the leopard by another lion (of the same weight as the lion) would add Replacing the leopard by another lion (of the same weight as the lion) would add 90

90 kg, and replakg, and replacincing the g the tigtiger by anotheer by another lion woulr lion would add 50d add 50 kg. kg. TheThen 3 n 3 liolions weins weighgh 310

310 + 90 + 90 + 50 = 450+ 50 = 450 kg kg and 1 and 1 liolion n weweighighs s 450450÷÷3 = 1503 = 150 kgkg,,

hence (B). hence (B).

Alternative 2 Alternative 2

If the lion wei

If the lion weighs 100ghs 100 kg, then the leopkg, then the leopard weiard weighs 10ghs 10 kg and the tiger 50kg and the tiger 50 kg for a kg for a totatotall of 160

of 160 kgkg. . ThThis is 150is is 150 kg too ligkg too lighht. t. AdAddiding 150ng 150÷÷3 = 50 kg to 3 = 50kg to eaceach weighh weight keeps thet keeps the

diﬀ

diﬀerenerences in ces in weweighight t the same. the same. So the So the liolion weighn weighs 150s 150 kg,kg,

hence (B). hence (B).

23.

23._{(Also J15, I15)}_{(Also J15, I15)}

As the number of points per event is 6 and the total number of points gained is As the number of points per event is 6 and the total number of points gained is 8 +

8 + 11 11 + 5 = + 5 = 24, there must hav24, there must have e been 4 been 4 evevenents. ts. As BettAs Betty y has 11 has 11 poinpoints she ts she mumustst have 3 ﬁrst places and one second place. Cathy could not have won an event, or her have 3 ﬁrst places and one second place. Cathy could not have won an event, or her score would be 6 or more, so she must have just one second place. So Adrienne must score would be 6 or more, so she must have just one second place. So Adrienne must have two second places,

have two second places,

hence (C). hence (C).

24.

24._{Arrange the coins as follows:}_{Arrange the coins as follows:}

A B B CC

JJanane e 55c c cocoiinns s 110c 0c ccooiinns s 550c 0c cocoiinnss T

Tom om 110c 0c ccooiinns s 220c 0c ccooiinns s 55c c cocoiinnss

Suppose Jane has just one 50c coin, so that Tom has one 5c coin and Jane has 45c Suppose Jane has just one 50c coin, so that Tom has one 5c coin and Jane has 45c more than Tom in column C.

more than Tom in column C.

Tom and Jane have the same amount, so than in columns A and B, Tom has 45c Tom and Jane have the same amount, so than in columns A and B, Tom has 45c more than Jane.

more than Jane. But for But for eaceach h coin in columncoin in columns s A and A and B, TB, Tom’om’s s coicoin n is wortis worth h twtwiceice Ja

Janene’s coin’s coin, , so Tso Tom has 90c om has 90c and Janand Jane e has 45chas 45c. . The fewThe fewesest t cocoinins s fofor r ththis is is is 1 1 coicoinn each in column A and 4 coins each in column B. Then they have 6 coins each.

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If Jane has more than one 50c coin, the diﬀerence in columns A and B will be 90c If Jane has more than one 50c coin, the diﬀerence in columns A and B will be 90c or more, whic

or more, which h requrequires more than 4 ires more than 4 coicoins in ns in thesthese e colucolumnsmns. . Then they havThen they have e moremore than 6 coins each.

than 6 coins each.

So the smallest number of coins they can each have is 6, when Jane has 1

So the smallest number of coins they can each have is 6, when Jane has 1×× 50c +50c +

44××10c + 110c + 1××5c and Tom has 15c and Tom has 1××5c + 45c + 4××20c + 120c + 1××10c,10c,

hence (D). hence (D).

25.

25. Alternative 1 Alternative 1

The ﬁnal mixture will have 9 litres of cordial out of 18 litres. The ﬁnal mixture will have 9 litres of cordial out of 18 litres.

The amount of the mixture from Jar A can vary from 0 litres to 18 litres, so we The amount of the mixture from Jar A can vary from 0 litres to 18 litres, so we try increasing amounts from Jar A and calculate the amount of 100% cordial in the try increasing amounts from Jar A and calculate the amount of 100% cordial in the mixture.

mixture. L

Liittrrees s ffrroom m JJaar r A A 0 0 1 1 2 2 33 ·· ·· ··

Liittrrees s ffrroom m JJaar r B B 118 8 117 7 116 6 1155 ·· ·· ··

Liittrrees s oof f ccoorrddiiaal l ffrroom m JJaar r A A 0 0 00..3 3 00..6 6 00..99 L

Liitrtres es oof f cocorrdidial al ffrrom om JJaar r B B 1010..8 8 1010..2 2 99..6 6 99..00 Li

Litrtres es of of cocordrdiial al iin n mimixtxturure e 1010.8 .8 1010.5 .5 1010.2 .2 9.9.99 ·· ·· ·· 9.09.0

For every additional litre from Jar A there is 0.3 litres less cordial in the mixture, For every additional litre from Jar A there is 0.3 litres less cordial in the mixture, which is 0.6 additional litres from Jar B and 0.3 fewer litres from Jar A.

which is 0.6 additional litres from Jar B and 0.3 fewer litres from Jar A.

Following this pattern, there will be 9 litres of cordial when there are 6 litres from Following this pattern, there will be 9 litres of cordial when there are 6 litres from Jar A and 12 litres from Jar B,

Jar A and 12 litres from Jar B,

hence (E). hence (E).

Note:

Note:_{Checking this, Jar A contributes 1.8 litres of cordial and Jar B contributes 7.2}_{Checking this, Jar A contributes 1.8 litres of cordial and Jar B contributes 7.2}

litres of cordial. litres of cordial.

Alternative 2 Alternative 2

Let the amount from Jar A be

Let the amount from Jar A be xx, and the amount from Jar B be, and the amount from Jar B be yy. Then. Then

x + + y y == 1188 ssoo xx = = 1818−−yy

00..33xx + 0 + 0..66yy = = 00..55××118 8 sso o 33xx + 6 + 6yy = = 990 0 aanndd xx = = 3030−−22yy

Therefore 18

Therefore 18−−yy = = 3030−−22yy, which has solution, which has solution yy = 12, and then = 12, and then xx = = 6,6,

hence (E). hence (E).

26.

26. Alternative 1 Alternative 1

Add

Adding 11 + 17 ing 11 + 17 + 22 = + 22 = 50 50 inclincludes everudes every y hat twichat twice. e. FFor or examexampleple, , QiaQiang’ng’s s hat is hat is inin both Rory’s and Sophia’s totals.

both Rory’s and Sophia’s totals. Th

Thereerefofore the re the tottotal of al of alall l ththree hats is 25. ree hats is 25. ThThen the en the perperson whoson whose totase total l is 11 is 11 hahass 25

25−−11 = 14 on their hat. The other two hats are 2511 = 14 on their hat. The other two hats are 25−−17 = 8 and 2517 = 8 and 25−−22 = 3.22 = 3.

So the three numbers are 3

So the three numbers are 3,,8 and 14,8 and 14,

hence (14). hence (14).

Alternative 2 Alternative 2

Let the numbers be

Let the numbers be a,a, bb andand cc. Then. Then aa + + b b = 11, = 11, bb + + c c = 17 and = 17 and aa + + c c = 22 = 22. . ThThenen 22aa + + 22bb + + 22cc = 50, therefore = 50, therefore aa + + b b + + c c = 25. Thi = 25. This gives givess aa = = 88,, bb = = 33,, cc = 14 so the = 14 so the largest number is 14,

largest number is 14,

hence (14). hence (14).

2016 AMC – Upper Primary Solutions 2016 AMC – Upper Primary Solutions 42

27.

27. Alternative 1 Alternative 1

We can test factors by division, ﬁnding each factor’s partner until we see a factor We can test factors by division, ﬁnding each factor’s partner until we see a factor that has occurred as a partner already.

that has occurred as a partner already. Factor Factor 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 9 8 9 1100 P Paarrttnneer r 88440 0 44220 0 22880 0 22110 0 11668 8 11440 0 11220 0 11005 5 – – 8844 F Faaccttoor r 111 1 112 2 113 3 114 4 115 5 116 6 117 7 118 8 119 29 200 P Paarrttnneer r — — 770 0 — — 660 0 556 6 — — — — — — — — 4422 F Faaccttoor r 221 1 222 2 223 3 224 4 225 5 226 6 227 7 228 8 229 39 300 P Paarrttnneer r 440 0 — — — — 335 5 — — — — — — 330 0 — — 2288 From this, there are 16 pairs of factors, giving 32 factors,

From this, there are 16 pairs of factors, giving 32 factors,

hence (32). hence (32).

Alternative 2 Alternative 2

Factorised into primes, 840 = 2 Factorised into primes, 840 = 233

×33××55××7. Then every factor of 840 can be found7. Then every factor of 840 can be found

by multiplying together: by multiplying together:

•

• a factor of 8 (1, 2, 4 or 8) a factor of 8 (1, 2, 4 or 8) •

• a factor of 3 (1 or 3) a factor of 3 (1 or 3) •

• a factor of 5 (1 or 5) a factor of 5 (1 or 5) •

• a factor of 7 (1 or 7) a factor of 7 (1 or 7)

This means there are 4

This means there are 4××22××22××2 = 32 possible factors of 840,2 = 32 possible factors of 840,

hence (32). hence (32).

28.

28._{(Also MP29)}_{(Also MP29)}

The ﬁrst cube

The ﬁrst cube useuses s 12 matc12 matches, then each subshes, then each subsequeequent cube uses 8 nt cube uses 8 matcmatcheshes. . SinSincece 2016

2016−−12 = 2004 and 200412 = 2004 and 2004÷÷8 = 2508 = 250r4, ther4, there are 1re are 1 ++ 250 = 2250 = 251 cubes mad51 cubes made, with 4e, with 4

matches left over, matches left over,

hence (251). hence (251).

29.

29._{(Also J27, I24)}_{(Also J27, I24)}

There are three cases for how the triangles are coloured: There are three cases for how the triangles are coloured:

(i) All three sides are the same colour, with 5 possibilities. (i) All three sides are the same colour, with 5 possibilities. (ii) Two sides are the same and one side is diﬀerent, with 5

(ii) Two sides are the same and one side is diﬀerent, with 5××4 = 20 possibilities.4 = 20 possibilities.

(iii) All three sides are diﬀerent, with

(iii) All three sides are diﬀerent, with 55××₆44₆××33 = = 10 10 possibilpossibilities.ities. So there are 5 + 20 + 10 = 35 possibilities in all,

So there are 5 + 20 + 10 = 35 possibilities in all,

hence (35). hence (35).

Note:

Note:_{The calculation in (iii) relies on the observation that if we are choosing the}_{The calculation in (iii) relies on the observation that if we are choosing the}

sides in order, there are 5 possibilities for the ﬁrst side, then 4 for the second, then sides in order, there are 5 possibilities for the ﬁrst side, then 4 for the second, then 3

3 for the thirfor the third. d. HoHowewevver, in er, in thethese 5se 5 ××44 ×× 3 = 60 possibilities, each selection3 = 60 possibilities, each selection _xyz_xyz

will appear 6 times:

will appear 6 times: xyxyz,z, xzxzy,y, yxyxz,z, yzxyzx,, zxzxy,y, zyxzyx. . ThThis ideis idea a appappearears in s in the genthe generaerall

formula for formula for



m m



_{, the number of ways of choosing}_{, the number of ways of choosing}

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30.

30. _If_Ifmm is a cousin’s age last year, it is a factor of 1377 andis a cousin’s age last year, it is a factor of 1377 andmm+ 1 is a factor of 2016.+ 1 is a factor of 2016. When we factorise we get 2016 = 2

When we factorise we get 2016 = 255

×3322 ××7 and 1377 = 37 and 1377 = 344 ××17.17.

Checking possible factors of 1377: Checking possible factors of 1377:

m m 1 1 3 3 9 9 117 27 27 57 51 81 81 11 1553 43 4559 19 1337777 m m+ + 1 1 2 2 4 4 110 0 118 8 228 8 552 2 882 2 11554 4 44660 0 11337788 Factor of 2016?

In document AMC Solutions Book 2016 (Page 39-71)

Solutions – Upper Primary DivisionSolutions – Upper Primary Division

Solutions – Upper Primary Division

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