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?

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