IAS 2020 — CSAT Maths

Correct Option

50

To determine the remainder when $51 \times 27 \times 35 \times 62 \times 75$51×27×35×62×75 is divided by 100, we need to evaluate the product modulo 100. The number 100 can be factored as $4 \times 25$4×25.

Let the given product be $P = 51 \times 27 \times 35 \times 62 \times 75$P=51×27×35×62×75.

We can express some of the factors in terms of their prime components or components related to 100:

• $35 = 5 \times 7$35=5×7
• $62 = 2 \times 31$62=2×31
• $75 = 3 \times 25$75=3×25

Substitute these into the product expression:

$P = 51 \times 27 \times (5 \times 7) \times (2 \times 31) \times (3 \times 25)$P=51×27×(5×7)×(2×31)×(3×25)

Rearrange the terms to group factors that form multiples of 100 or 50:

$P = (5 \times 2 \times 25) \times (51 \times 27 \times 7 \times 31 \times 3)$P=(5×2×25)×(51×27×7×31×3)

$P = (10 \times 25) \times (51 \times 27 \times 7 \times 31 \times 3)$P=(10×25)×(51×27×7×31×3)

$P = 250 \times (51 \times 27 \times 7 \times 31 \times 3)$P=250×(51×27×7×31×3)

Let $K = 51 \times 27 \times 7 \times 31 \times 3$K=51×27×7×31×3. All the individual factors (51, 27, 7, 31, 3) are odd numbers. The product of any number of odd integers is always an odd integer. Therefore, K is an odd integer.

Now, we need to find the remainder of $250 \times K$250×K when divided by 100. We know that $250 \equiv 50 \pmod{100}$250≡50(mod100).

So, the product $P \equiv 50 \times K \pmod{100}$P≡50×K(mod100).

Since K is an odd number, it can be written in the form $2m + 1$2m+1 for some integer m.

$P \equiv 50 \times (2m + 1) \pmod{100}$P≡50×(2m+1)(mod100)

$P \equiv (100m + 50) \pmod{100}$P≡(100m+50)(mod100)

$P \equiv 50 \pmod{100}$P≡50(mod100)

Thus, the remainder when the given product is divided by 100 is 50.

Incorrect Options:

The options 25, 5, and 1 are incorrect because the rigorous application of modular arithmetic, as demonstrated above, yields a remainder of 50. Any other remainder would contradict the derived result. For instance, a remainder of 25 would imply that the product is congruent to 25 modulo 100, which is not the case. Similarly, remainders of 5 or 1 are inconsistent with the calculated value of 50.