Are you Smarter than a First Grader? (3)

1st grader 2

Q. AB – CD = EF; EF + GH = PPP. A, B, C, D, E, F, G, H, and P are unique digits different from each other. What’s A, B, C, D, E, F, G, H, and P?

A. For sure you can’t solve 9 unknowns with only 2 equations with algebra alone. But you can use logic and trial and error. One method is as follows:

1. First, the sum of any two 2-digit numbers can never be greater than or equal to 200. Hence P must be less than 2. It cannot be 0 because otherwise E, F, G, and H must all be 0, which is not allowed by the problem definition. Therefore P must be 1.
2. Secondly, since EF + GH = 111, therefore F + H must = 11, carrying 1 to E + G and E + G must = 10, thus making it 11.
3. E + G =10 possibilities include 8+2, 7+3, 6+4, 4+6, 3+7, 2+8. It cannot be 9+1 or 1+9, as 1 = P already. Nor can it be 5+5, as each alphabet can be used only once.
4. F + H = 11 possibilities include 9+2, 8+3, 7+4, 6+5, 5+6, 4+7, 3+8, 2+9.
5. Since each digit appears only once, each EG combination would preclude those FH combinations in which the EG digits are already used e.g. for E+G as 8+2 precludes the four F+H combinations of 9+2, 8+3, 3+8 and 2+9.
6. As there are six EG and eight FH combinations, four of which are precluded by each set of EG digits, there is a total of 6 X (8-4) = 24 possibilities, which are listed as follows:
a. 87 + 24 = 111, 86 + 25 = 111, 85 + 26 = 111, 84 + 27 = 111;
b. 79 + 32 = 111, 76 + 35 = 111, 75 + 36 = 111, 72 + 39 = 111;
c. 69 + 42 = 111, 68 + 43 = 111, 63 + 48 = 111, 62 + 49 = 111;
d. 49 + 62 = 111, 48 + 63 = 111, 43 + 68 = 111, 42 + 69 = 111;
e. 39 + 72 = 111, 36 + 75 = 111, 35 + 76 = 111, 32 + 79 = 111;
f. 27 + 84 = 111, 26 + 85 = 111, 25 + 86 = 111, 24 + 87 = 111.
7. Thirdly AB – CD = EF. Since A to H and P are all digits between 0 to 9, A must be > C and A > E as negative numbers are not allowed.
8. For each of the EF + GH = 111 listed in 6a to 6f, only the digits not already used by E, F, G, and H can be assigned to A, B, C and D to fit AB – CD = EF. For example, in 6a for 87 + 24 = 111, only 9, 6, 5, 3 and 0 remain to be used. In this particular case, only 93 – 06 would satisfy AB – CD = 87.
9. One can go through all the 24 possibilities in 6a to 6f, and by trial and error find out all the solutions that would fit each equation. Note that not all possibilities have a valid solution. For example, for 6f, while there is a solution for 27 + 84 = 111 (90 – 63 = 27), no combination of 9, 6, 5, 3 and 0 would satisfy AB – CD = 24 such that 24 + 87 = 111. Thus there are multiple solutions for AB + CD = EF, EF + GH = PPP. Perhaps you can write an algorithm to test all combinations.

Now, given the problem’s complexity, do you expect the average first grader to solve it? Or has the zealous examiner gone way overboard? Even though most people are not into mathematics and few would read this post, I wrote it not only because I enjoy mental challenges, but because there are similarities between this and bible study. As I said before, sometimes you need good detective skills to interpret the Bible properly.

First there is observation, noting all the details and the relationship between them. Some people just notice the prima facie evidence and jump right to conclusion. That would not do. Then there is interpretation. Some interpretations are possible until precluded by other observations not yet taken into account. You need to consider all biblical evidence and accept only the solution(s) that harmonize everything, because God does not make mistakes. Finally, sometimes you just have to accept the fact that there are no solutions this side of heaven, because God had chosen not to reveal everything to us. The mysteries belong to God. So come to the Bible with a keen but humble mind, ready to apply everything you learn and you will have a marvelous time learning from the Word.

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