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# Solution

**1.** We are given that PARTS × 4 = STRAP. So, a five-digit number multiplied by 4 gives a five-digit number.

PARTS

× 4

STRAP

**2.** Now, if PARTS were 25000, then 25000 × 4 = 100000, which is a six-digit number. Therefore, the maximum value that PARTS can take is 24999
because when you multiply 24999 by 4, you get 99996, which is the largest five-digit number possible in this case. So, PARTS has to be less than 24999
(it cannot be equal to 24999 because all the digits have to be different and none of them is zero). This means, P can either be 1 or 2.
The first step in multiplying PARTS by 4 is to multiply S by 4; the result of this product should give P in the unit's place. S can take values from 1 to 9,
but none of these values yields 1 in the unit's place when multiplied by 4. So, P cannot be 1; it has to be 2. So, we get our first digit, **P = 2**.

**3.** The product of S and 4 yields 2 in the unit's place only when S = 3 or 8 (4 × 3 = 12; 4 × 8 = 32). In the last step of the multiplication
process, you have to multiply P by 4. Now, P = 2. So, P × 2 = 8. If there is a digit carried from the previous step (A × 4), it can only be 1, 2, or 3
(the maximum value of a digit is 9; when multiplied by 4, you get 36. So, the maximum value of the carried digit is 3). If the carried digit is 2 or 3, then,
(P × 2 + carried digit) will be either 10 or 11 and the result will be a six-digit number. But the result is a five-digit number. So, the carried digit cannot
be 2 or 3. It can be 1. In that case, P × 2 + 1 = 9. But P × 2 + carried digit = S, according to the final result. So, S = 9. We have seen that S can only
be either 3 or 8. So, we can conclude that there was no carried digit, and **S = 8**.

So, our number is 2ART8. We have still to find A, R, and T.

**4.** At the time of finding S, we established that the product of A and 4 did not produce a carried digit. So, A × 4 is a single-digit number. If A were 3
or greater than 3, then the product will be a double-digit number. So, A has to be less than 3. It cannot be zero (none of the digits is zero); so, it can be either 1 or 2.
But it cannot be 2 because we already have P = 2 (all the digits are distinct). So, **A = 1**.

**5.** So, now the number becomes 21RT8. And, 21RT8 × 4 should be 8TR12. Let's write this as:

21RT8

× 4

8TR12

**6.**So, we begin by multiplying 8 by 4. We get 32. We carry forward 3 and add it to the product of T and 4. So, we have 4T + 3. Now, 4T + 3 should have 1 in
the one's place. This is possible only when T = 2 (4T + 3 = 11) or when T = 7 (4T + 3 = 31). But T cannot be 2 because we already have A = 2. So, **T = 7**.

**7.** So, now, we have the number as 21R78.

**8.** So, 4T + 3 = (4 ×7) + 3 = 31. Here, we have 1 in the one's place and 3 is carried forward and added to the product of 4 and R. So, in the next step, we
have (R × 4) + 3. The result should have R in the unit's place. This is possible only when R = 9 because (9 × 4) + 3 = 39. So, **R = 9**.

**9.** Thus, **the required five-digit number is 21978**. When you multiply 21978 by 4, you get 87912, which is the original number reversed. Hope I have done it right.
I am sure there must be a simpler way of solving it.

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- Union Public Service Commission - www.upsc.gov.in
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