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Key Stage 2 - Team Contest
1. Let each of the letters D, A, V, O, M, T, H and S represent a distinct digit from 0 to 9 so that DAVAO and MATHS are 5-digit numbers and it satisfies:
D A V A O
＋ D A V A O M A T H S
Find the sum of all possible values for T.【Submitted by Bulgaria_FPMG】 【Solution】
From the thousands digits, it is clear that A is 0 or 9.
If A=0, then H =1 from the tens digits and hence we have: (i) O+ =O HS>10, i.e. O≥6 and S =2, 4, 6 or 8.
(ii) V + = <V T 10, i.e. 2≤ ≤V 4 and T =4, 6 or 8. (iii) D+ =D M <10, i.e. 2≤ ≤D 4 and M =4, 6 or 8.
If D=2, then M = 4 and hence V =3. So T =6. Thus, S =8 and O=9. We get a solution:
2 0 3 0 9 ＋ 2 0 3 0 9 4 0 6 1 8 If D=3, then M =6.
As V =2, then T =4 and hence, S =8. So O=9. We get a solution: 3 0 2 0 9
＋ 3 0 2 0 9 6 0 4 1 8
As V =4, then T =8 and hence, S = 2. So O=6, which is impossible. If D=4, then M =8 and hence, T =6. So V =3. Thus, S =2 and we can conclude that O=1, which is impossible.
If A=9, then H =8 and O+ = <O S 10. Since H =8, we have S =2, 4 or 6 and 1≤ ≤O 3. We also have:
(ii) D+ + =D 1 M <10. Since A=9, we have 1≤ ≤D 3 and M =3, 5 or 7. If V =5, then T =1.
As M =3, then D=1, which is impossible.
As M =7, then D=3 and hence, O=2. So S =4. We get a solution: 3 9 5 9 2
＋ 3 9 5 9 2 7 9 1 8 4 If V =6, then T =3.
As O=1, then S =2 and hence, D=0, which is impossible.
As O=2, then S =4 and hence, D=1. So M =3, which is impossible. If V =7, then T =5 and hence, M =3. So D=1 and O=2, S =4. We get a solution:
1 9 7 9 2 ＋ 1 9 7 9 2 3 9 5 8 4
Thus, there are four possible values for T, namely 6, 4, 1 and 5. So the answer is 16.
Answer: 16
2. The figure below shows a quadrilateral ABCD so that ∠ADC = ∠ABC= °90 and AD=DC. If the area of the quadrilateral ABCD is 196 cm2, find the distance from D to AB, in cm.【Submitted by Central Jury】
【Solution 1】
Let point E on AB and point F on the extension of BC so that DE//BC and DF//AB. (10 Marks) Since ∠ABC= °90 , ∠AED= ∠DFC = °90 . Observe that
90
ADE EDC CDF EDC
∠ + ∠ = ° = ∠ + ∠ , so ∠ADE = ∠CDF. Since AD=DC,
triangle ADE and CDF are congruent (10 Marks) and hence DE= DF (10 Marks). Thus, DEBF is a square and the area of DEBF is equal to the area of ABCD. Thus
14 DE = cm since DE2 =196 14= 2. (10 Marks) A B C D E F
【Solution 2】
Let point E on AB so that DE ⊥ AB. Since ∠ADC = ∠ABC= °90 , 360 90 90 180
DAE BCD
∠ + ∠ = ° − ° − ° = °. Thus as we rotate triangle DAE 90° counterclockwise to triangle DCF so that DA and DC coincides (10 Marks) , then F,
C and B are collinear. (10 Marks) So BEDF is a square and the area of BEDF is equal to the area of quadrilateral ABCD, DE2 =196 14= 2. Hence DE=14cm. (10 Marks)
Answer: 14 cm
3. Connect each small box on the top with its same letter on the bottom with paths that do not cross one another, nor leave the boundaries of the large box.
Connect each letter in the box on the top with the same letter on the bottom with paths not crossing over one another, nor paths not going outside the border . 【Submitted by Sri Lanka】
【Solution】
One of many possible ways is shown in the diagram below.
O
I
T
I
M
T
O
M
O
I
T
I
M
T
O
M
4. Fill in each □ with digit from the set {1, 2, 3, 4, 5, 6, 7, 8, 9} with no repetition into the following math operation:
1 1 1 1 1 1 M N + + = + + + □ □ □
□
□
□
, where M and N are relatively prime.
Find greatest possible value of M −N.【Submitted by Indonesia】 【Solution】 Simplifying, we get 1 1 1 1 1 1 1 1 1 c ef a b d a bc def d f c e f + + + = + + + + + + + +
From the above equation, we know a =1. To make
1 c bc+ greatest we choose c=9 and b=2. To make ef 1 def d f +
+ + greatest, we choose d =3, e=7 and f =8. So, the equation becomes
1 1 1 9 7 8 1 1 1 2 9 1 3 7 8 3 8 9 57 1 19 179 3401 1611 1083 3401 6095 3401 c ef a bc def d f + × + + + = + + + + + × + × × + + = + + + + = =
So, the value M – N = 6095 – 3401 = 2694 【Marking Scheme】
Be able to give the correct value of a, 5 marks Be able to give the correct value of b and c, 5 marks Be able to give the correct value of d, e and f, 10 marks Be able to give the correct value of M and N , 10 marks Be able to give the correct difference M – N, 10 marks
5. Form edges along the dotted lines to create shapes so that each circle is the symmetry center (each shape when rotated 180 degrees along the circle, the shape appears identical) of the enclosed area. An example is shown below. Complete the three challenges below.【Submitted by Indonesia】
Example :
Challenge
【Solution】
【Marking Scheme】
Each correct figure, 10 marks. 10 marks bonus for all 3 correct figures.
6. There are three 3-digit numbers ABC, BCD and CDE, where each different letter represents a different digit, so that ABC+ BCD+CDE =2017. Find the difference between the largest and smallest possible value of ABCDE .
【Solution】
Observe that all of A, B and C are not equal to 0. Now we have A B C
B C D ＋ C D E 2 0 1 7
Since sum of three different digits are not greater than 9 8+ + =7 24 and not less than 0 1 2+ + =3, there are following 4 cases:
(i) 7 11 19 C D E B C D A B C + + = + + = + + = ; (ii) 7 21 18 C D E B C D A B C + + = + + = + + = ; (iii) 17 10 19 C D E B C D A B C + + = + + = + + = ; (iv) 17 20 18 C D E B C D A B C + + = + + = + + = In case (i), B= +E 4 and A= + ≥D 8 8.
In case (ii), B= +E 14, this is in contradiction with B<10.
In case (iii), E= +B 7 and A= +D 9. So A=9, D=0 and hence C+ =E 17, this is in contradiction with C+ ≤ + =E 7 8 15.
In case (iv), B= +E 3 and D= +A 2. So A≤7. Thus only case (i) and (iv) can work.
To minimize ABCDE , we consider case (iv) first. If A=1, then D=3 and we can get B+ =C 17 and C + =E 14. Thus we can take B=8, C =9 and E=5. So the smallest possible value of ABCDE is 18935.
To maximize ABCDE , we consider case (i) first. If A=9, then D=1 and hence 10
B+ =C and C+ =E 6. Thus we can take B=8, C =2 and E =4. So the largest possible value of ABCDE is 98214.
So the difference between the largest and the smallest possible value of ABCDE is 98214 18935− =79279.
Answer: 79279 【Marking Scheme】
Find the largest possible value of ABCDE , 15 marks Find the smallest possible value of ABCDE , 15 marks Get the correct answer, 10 marks.
7. Andy wants to put some tokens (can put 0 token) on each of the unit squares of a
3 3× board, so that in any 2 2× subboard, the sum of the number of tokens is a prime number and all of those prime numbers are different. What is the least possible number of tokens that Andy should put on the board? Show one example. 【Submitted by Central Jury】
【Solution】
Let 2≤ p1 < p2 < p3 < p4 ≤7 be the number of tokens in the 4 squares of 2 2× subboards. There is a 2 2× subboard with at least 7 tokens, therefore, Andy puts at least seven tokens. These is a possible way to put the tokens:
Answer: 7
8. The figure shows a triangle ABC. D is the midpoint of BC and E lies on AC such that
AE : EC = 3 : 2. If F is a point of AB such that the area of triangle DEF is three times the area of triangle BDF, find the ratio of AF : FB. 【Submitted by Central Jury】
【Solution】
Connect CF and AD and let the area of triangle ABC be 10. Then the area of triangle ADC is 5 and
hence the area of triangle CDE is 2. Thus the area of quadrilateral ABDE is 8. Suppose AF : FB = a : b.Then the area of triangle AFC is 10a
a+b and hence the area
of triangle AEF is 6a
a+b.And the area of triangle BFC is
10b
a+b and hence the area
of triangle BDF is 5b
a+b. Thus the area of triangle DEF is
15b
a+b and hence the area
of quadrilateral BDEF is 20b
a+b. Since quadrilateral ABDE is formed by triangle AEF
and quadrilateral BDEF, we have 20b 6a 8
a+b + a+b = , i.e. 20b+6a =8a+8b, or
6
a= b. So AF : FB = a : b = 6：1.
Answer: 6：1 【Marking Scheme】
This problem has many different potential solutions and approaches
Expressing the area of triangle DEF in terms of AF:FB up to 10 Marks
(depending of progress)
Expressing the area of triangle BDF in terms of AF:FB up to 10 Marks
(depending on progress)
Solving for AF:FB based on the 2 previous relationships up to 10 Marks
(depending on progress)
Guessing the ration 10 Marks
0 0 0 0 2 1 0 3 1 A B C D F E
9. Use each of the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 once to form a 2-digit number, a 3-digit number and a 4-digit number, respectively, so that when the first two numbers are multiplied together, the result would be the third number. For example, 12 483× =5796.
Aside from the example given above, list down all possible combinations.
【Submitted by Thailand】
【Solution】
There are six more possible combinations (as shown below), aside from the example given in the problem.
18 297 5346 27 198 5346 28 157 4396 39 186 7254 48 159 7632 42 138 5796 × = × = × = × = × = × = 【Marking Scheme】
Each correct combination, 6 marks. 4 marks bonus for all 6 correct figures.
10. The picture shows a part of the Balkan peninsula. In how many ways can we color each region with one of 4 colors, so that every 2 neighboring regions are colored with different colors? 【Submitted by Bulgaria_FPMG】
【Solution 1】
WLOG let BG be colored in color a, RU in color b and SRB in color c which we can do in 4 3 2× × =24 ways. (10 Marks) Depending on the ways we color MK, we can color GR, TR, BS as following, there are 11 ways. (20 Marks)
MK GR TR BS b c b c 1 d 2 d c 3 d b c 4 d 5 c d 6 d b c d 7 d c 8 c b c 9 d 10 d c 11
Finally, the total number is: 24(3 2 5 1)× + × =264 ways. 【Solution 2】
Let’s use an approach where we fix the color for a specific country and find out how many choices we have for the adjacent countries. We will use the order RU
SRB BG MK GR TR BS
There are 4 colors available for RU. Choose one of them (say A). There are 3 colors now available for SRB. Choose B.
There are 2 colors available now for BG (as A and B are taken by adjacent countries). Say we choose C.
So far we have 4 x 3 x 2 = 24 combinations for (A,B,C)
For MK, GR and TR we have 2 options for coloring each once we have fixed the previous country, namely:
MK two colors as it cannot be B and C
GR there is a choice of two colors as it cannot be the same color as BG and cannot be the same color as MK. So 2 choices for GR.
TR Once BG and GR are fixed we have 2 options for TR. Therefore for MK, GR and TR we have 8 combinations.
D,A,B D,A,D D,B,A D,B,D A,B,A A,B,D A,D,A A,D,B
Now we have to find how many options we have for BS. If TR, BG and RU are all in different colors then we have only one option for BS. However, if TR and RU have the same color then we have 2 options for BS.
Out of the 8 options above in 3 options TR and RU are in the same color (A) and in 5 options TR and RU are in different color.
Therefore, the total options are 24 x (5 x 1+ 3 x 2)=264 【Marking Scheme】
Choosing 3 countries and fixing their colors 5 pts
Proper reasoning for the rest of the countries based on restrictions of already fixed colors up to 10 pts (here partial credits will be given based on the progress)
Finding the correct answer (even with guessing) 5 pts
Providing a working counting strategy (different from the solution) up to 15 pts (based on the progress)
