Find the last 2 digits of the following number-75 - 35 - 47 - 63 - 71 - 87 - 82

The correct option is A 50[75 × 35 × 47 × 63 × 71 × 87 × 82]/100=3 × 35 × 47 × 63 × 71 × 87 × 41/2→ remainder =1. Hence, required remainder =1 × 50 =50. ...

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Last Two Digits of a Number

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Question

Find the last 2 digits of the following number:
$75 \times 35 \times 47 \times 63 \times 71 \times 87 \times 82$

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Solution

The correct option is A 50
$\frac{[75 \times 35 \times 47 \times 63 \times 71 \times 87 \times 82]}{100} = \frac{3 \times 35 \times 47 \times 63 \times 71 \times 87 \times 41}{2} \to$ remainder =1.
Hence, required remainder = $1 \times 50 = 50.$

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Find the last 2 digits of the following number: 75×35×47×63×71×87×82

The correct option is A 50[75×35×47×63×71×87×82]100=3×35×47×63×71×87×412→ remainder =1. Hence, required remainder =1×50=50.

Find the last 2 digits of the following number:
$75 \times 35 \times 47 \times 63 \times 71 \times 87 \times 82$

The correct option is A 50
$\frac{[75 \times 35 \times 47 \times 63 \times 71 \times 87 \times 82]}{100} = \frac{3 \times 35 \times 47 \times 63 \times 71 \times 87 \times 41}{2} \to$ remainder =1.
Hence, required remainder = $1 \times 50 = 50.$