Are you Smarter than a First Grader? (4)

1st grader 5

My wife joined the reunion of her high school classmates, and they gave her the following problem:
1111 = 0 | 5555 = 0 | 2222 = 0 | 8193 = 3 | 7662 = 2
8096 = 5 | 9313 = 1 | 4398 = 3 | 0000 = 4 | 9475 = 1
6666 = 4 | 9038 = 4 | 7172 = 0 | 3148 = 2 | 2889 = 5
1964 = ?

They did not say what grade this is, but supposedly it can be solved by a first grader. And unlike the previous two problems which require logic and deduction, this one is really simple, so simple that we who are so complicated often overlook the solution since it is so obvious. What does 1964 equal to? Can you guess?

The answer is 2! Why? Need a hint? Don’t try to see if there is a secret code hidden in those 15 four-digit numbers. I was stumped when I tried to discern any relationship between them, or between the digits that made them up. In vain did I try to sum the digits in each number, or do all sorts of mental gymnastics to “massage” the data! There is no need for any of that. Mt 11:25 At that time Jesus said, “I praise You, Father, Lord of heaven and earth, that You have hidden these things from the wise and intelligent and have revealed them to infants. (also Lk 10:21) We who consider ourselves wise are too intelligent for our own good, and the simplicity of things just eludes us.

Still clueless? What shape do little children like, that parents feel safe about? “o”! Count how many “o”s there are in each of the numbers and you’ll get your answer! Now do you see? I have two degrees, with distinction, but I hadn’t got a clue! Humbling isn’t it? Reminds me of Jas 4:6 But He gives a greater grace. Therefore it says, “God is opposed to the proud, but gives grace to the humble.” (also 1 Pet 5:5).

Mt 18:3 and said, “Truly I say to you, unless you are converted and become like children, you will not enter the kingdom of heaven. (also Mk 10:15; Lk 18:17) Hope we all have the simple faith of children totally dependent on their parents.

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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.

Are You Smarter than a First Grader? (2)

1st grader 1

My brother sent me two math exam questions, supposedly grade 1. Let’s see if you are smarter than a first grader.

(Q1) ABCD X 9 = DCBA. Each of the 4 alphabets A, B, C, and D represents a different digit. What’s A, B, C and D?
(Q2) AB – CD = EF, EF + GH = PPP. Each of the 9 alphabets represents a different digit. What are A, B, C, D, E, F, G, H and P?

As far as I can remember, algebra wasn’t taught until high school. But that was over half a century ago. Maybe today’s kids are much smarter. But for Q1, there are 4 unknowns and only one equation. For Q2, there are 9 unknowns and only 2 equations. You need as many equations as the number of unknowns, so how do you solve the problem algebraically? Algebra won’t work in this instance. But a little logic might.

In Q1, the reasoning could be as follows:
* Each of A, B, C, D could be any unique number between 0 to 9.
* For a 4 digit number ABCD to be multiplied by 9 and still remain to be 4 digits, A can only be 1. It cannot be >1 as then the product would be 5 digits, not 4. A cannot be 0 either as then BCD X 9 could not produce a 4-digit number ending in 0.
* So for now we have 1BCD X 9 = DCB1. But the only multiple of 9 that produces a number ending in 1 is 9 i.e. 81. Therefore D = 9.
* Now we have 1BC9 X 9 = 9CB1, and B, C, could be 0, or a number between 2 to 8.
* But if B is any number between 2 to 8, when multiplied by 9 it would carry over from “hundreds” to “thousands”, which is not permissible by the previous steps. Hence B can only be 0.
* Now we have 10C9 X 9 = 9C01, with C being a number between 2 to 8.
* By trial and error, the only number that fits is 1089 X 9 = 9801. Therefore C = 8.

That’s a fair bit of logical and arithmetical thinking. Do we expect that from a first grader, a six-year old? And if you think that’s hard, try Q2. Any math teacher out there who can offer some help?