Page 144 - NUMINO Challenge_B1
P. 144
Type 2-2 Alphametics (2) p.20~p.21 Creative Thinking p.22~p.23
1 Given that no number is carried 1 B CD 1 Since BBB 111 B 3 37 B, ABC 37 B.
over when A is multiplied by 9, 9
A 1. When B 1, 2, 3, , the results of 37 B are
DC B 1
2 D 9, the only number that can shown below.
be multiplied by 9 to result in a 1BC9
1 in the ones place. 9 B123456789
37 B 37 74 111 148 185 222 259 296 333
3 B 9 C. Given that no number 9CB1
is carried over when B is Use B 4 to satisfy the equation ABC 37 B.
multiplied by 9, B 0. 10C9 Since ABC 148, A B C 13.
9
4 Remember the 8 that is carried 2 B is the digit in the ones place of (the product
over as a result of 9 9 81. 9C01
C 8, the only number that will of) B B, so it can either be 1, 5, or 6. However
result in a 0 in the ones place. 1089
9 B 1, because AB B 3CB.
9801 If B is 5, A is either 6 or 7. Because A 7 leads to
C 7, use A 6 and C 2 to satisfy the equation
AB B 3CB. However, when you multiply 65 by
25, it is not satisfied AB C 1BD. 56
So B 6 and A 5. 56 6 336. 36
Therefore, C 3, and EFGB is 336
Problem solving 2016. 168
2016
1 Read the Key Point. The letter 'a' can either be 3 In the ones place, 'b' must be equal aba
ca
1 or 2. to 0, because a a 0. Since the ab
greater number has only 3 digits and
If a 1, the digit in the ones place in the the difference has only 2 digits, 'a' 101
can only be 1. c1
answer must also be 1. But 1 can never be Therefore, a 1, b 0, c 9. 10
the sum of an even amount of numbers, so
'a' cannot be 1.
If a 2, d 8. The sum of 2bc8
four 0's or 1's will not have 2bc8
any carry-on numbers. So 'b' 2bc8 4 Because 0 0 0, the only digit that can be D is
is 0 or 1. The digit in the 2bc8 0. The sum of two BAD's will result in an answer
over 1000. So, G 1. Calculate different possible
ones place of the sum 8cb2 sums doubling the same number. Three
possibilities are 720, 830, and 940. The numbers
c c c c 3 equals b, so that make up GOOD are 720 720 1440,
830 830 1660, and 940 940 1880.
b 1 and c 7. Therefore, abcd is 2178.
2 To solve the equation AB A CCC, 37
3
CCC can only be 111. Therefore,
11 1
A 3, B 7, and C 1.
Answer Key
1 Given that no number is carried 1 B CD 1 Since BBB 111 B 3 37 B, ABC 37 B.
over when A is multiplied by 9, 9
A 1. When B 1, 2, 3, , the results of 37 B are
DC B 1
2 D 9, the only number that can shown below.
be multiplied by 9 to result in a 1BC9
1 in the ones place. 9 B123456789
37 B 37 74 111 148 185 222 259 296 333
3 B 9 C. Given that no number 9CB1
is carried over when B is Use B 4 to satisfy the equation ABC 37 B.
multiplied by 9, B 0. 10C9 Since ABC 148, A B C 13.
9
4 Remember the 8 that is carried 2 B is the digit in the ones place of (the product
over as a result of 9 9 81. 9C01
C 8, the only number that will of) B B, so it can either be 1, 5, or 6. However
result in a 0 in the ones place. 1089
9 B 1, because AB B 3CB.
9801 If B is 5, A is either 6 or 7. Because A 7 leads to
C 7, use A 6 and C 2 to satisfy the equation
AB B 3CB. However, when you multiply 65 by
25, it is not satisfied AB C 1BD. 56
So B 6 and A 5. 56 6 336. 36
Therefore, C 3, and EFGB is 336
Problem solving 2016. 168
2016
1 Read the Key Point. The letter 'a' can either be 3 In the ones place, 'b' must be equal aba
ca
1 or 2. to 0, because a a 0. Since the ab
greater number has only 3 digits and
If a 1, the digit in the ones place in the the difference has only 2 digits, 'a' 101
can only be 1. c1
answer must also be 1. But 1 can never be Therefore, a 1, b 0, c 9. 10
the sum of an even amount of numbers, so
'a' cannot be 1.
If a 2, d 8. The sum of 2bc8
four 0's or 1's will not have 2bc8
any carry-on numbers. So 'b' 2bc8 4 Because 0 0 0, the only digit that can be D is
is 0 or 1. The digit in the 2bc8 0. The sum of two BAD's will result in an answer
over 1000. So, G 1. Calculate different possible
ones place of the sum 8cb2 sums doubling the same number. Three
possibilities are 720, 830, and 940. The numbers
c c c c 3 equals b, so that make up GOOD are 720 720 1440,
830 830 1660, and 940 940 1880.
b 1 and c 7. Therefore, abcd is 2178.
2 To solve the equation AB A CCC, 37
3
CCC can only be 111. Therefore,
11 1
A 3, B 7, and C 1.
Answer Key
