cleetus wrote:
In a four-digit number, the sum of the digits in the units place and the tens place is equal to the sum of the digits in the hundreds and the thousands places. The sum of the digits in the tens and hundreds place is twice the sum of the other two digits. If the sum of the digits of the number is more than 20, then the digits in the units place can be
A) 6
B) 7
C) 8
D) 5
E) 2
Let's call the number ABCD.
We're given the following information:
(1) A + B = C + D
(2) B + C = 2(A + D) = 2A + 2D
(3) A + B + C + D > 20
This isn't much to go on, so we should try to derive some more information from what we're given. First, keep in mind that A, B, C, and D are digits - numbers between 0 and 9 inclusive, except for A which can not be 0. Based on equation (2) we can guess that A and D will be smaller than B and C. In fact, since B + C <= 18, we know that A + D <= 9. If A + D = 9, then B and C would both be 9 and in order to fulfill equation (1) A and D would have to be 9/2, but this isn't a digit... so we can do even better A + D < 9. Finally, if we know A and D we can find B and C by combining a different form of equation (1) and equation (2)
D - A = B - C
2A + 2D = B + C
This is a lot better than where we started.
Since this is a "can be" question, we should proceed by elimination. We'll start with the largest answer first as this puts more restrictions on A, and once we have A and D we can find B and C.
Suppose D = 8. Then A = 0 - NO
Suppose D = 7. Then A = 1 and B + C = 16 and B - C = 6. This implies B = 11. NO
Suppose D = 6.
And A = 1. So, B + C = 14 and B - C = 5. This implies B = 19/2. NO
And A = 2. So, B + C = 16 and B - C = 4. This implies B = 10. NO
Suppose D = 5.
And A = 1. So, B + C = 12 and B - C = 4. This implies B = 8 and ABCD = 1845, but 1 + 8 + 4 + 5 = 18 < 20. NO
And A = 2. So, B + C = 14 and B - C = 3. This implies B = 17/2. NO
And A = 3. So, B + C = 16 and B - C = 2. This implies B = 9 and ABCD = 3975, and 3 + 9 + 7 + 5 = 24 > 20. YES
BenchPrepGURU