Question

The last two-digits in the multiplication 122×123×125×127×129 will be

• A. 20
• B. 50
• C. 30
• D. 40

Answer 1

The Correct Option is B

To find the last two digits of the multiplication \(122 \times 123 \times 125 \times 127 \times 129\), we focus on calculating this product modulo 100. Essentially, we seek to compute:

\( (122 \times 123 \times 125 \times 127 \times 129) \mod 100 \)

We can reduce each factor modulo 100:

• \(122 \equiv 22 \mod 100\)
• \(123 \equiv 23 \mod 100\)
• \(125 \equiv 25 \mod 100\)
• \(127 \equiv 27 \mod 100\)
• \(129 \equiv 29 \mod 100\)

Our aim is to calculate:

\((22 \times 23 \times 25 \times 27 \times 29) \mod 100\)

Notice that one of the factors, 125, ends in 5. Therefore, the entire product will end in zero. Now, examine if the product is a multiple of 100, which requires identifying if the product is divisible by both 4 and 25 (since \(100 = 4 \times 25\)).

We know:

• The product already contains a factor of 25 from 125.
• Now check divisibility by 4: Among the factors, 22, 23, 27, and 29 are not multiples of 4, but compound calculations show that the product modulo 4 impacts whether the final two digits are strictly zero or end in 50 instead of zero.

Combining these details carefully, without needing full expansions but by careful factor analysis:

The result sticks to options ending in 50 due to this arrangement analysis.

Thus, the final result is:

50