# Intelligent Math Puzzle – Challenge 88

A great *mathematical* and logic *puzzle* designed for those interested in school-level *Maths*. The answer is also given.

## Challenge:

In the figure above, A, B, C, and D represent different digits. What is the value of A × B + C + D?

**A-** 6

**B-** 12

**C-** 34

**D-** 40

**E-** 45

### The Absolute Best Book** to ****challenge your Smart Student!**

The correct answer is A.

A, B, C, and D represent different digits. The sum of three 3-digit numbers is a 4-digit number with 1 or 2 in the thousands place. Therefore, D = 1 or 2. And A is 3 or greater than 3. Why?

A + A + A = 12 or 12+1, Why?

Since, A is a digit, then 3A cannot be 13. 3A = 12 ⇒ A = 4

We also know that C + C + C = C or C + 10

C = 0 or 3C = C + 10 → 2C = 10 → C = 5

Therefore, C is 0 or 5.

Let’s assume D = 1. Then:

If C = 0 → B + B + B = 1 or B + B + B = 1 + 10

3B = 1 (This is not correct) or 3B = 1 + 10 (This is not correct either).

If C = 5 → B + B + B + 1 = D → D = 1 → 3B = 1 – 1= 0

or

B + B + B + 1 = D + 10 → 3B + 1 = 1 + 10 → B \(= \frac{10}{3}\)

B is an integer and cannot be \(\frac{10}{3}\).

In this situation, the only value of B is 0.

If A = 4, B = 0 and C = 5, then 405 + 405 + 405 = 1215 Bingo!

You can check out other values for A, B, C, and D. But, only the above values work.

A × B + C + D = 4 × 0 + 5 + 1 = 6

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