What is the least perfect square which leaves the remainder 1 when divided by 7 as well as by 11?

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Solution

Let the required least perfect square be x².

Then, by the given condition, (x²− 1) is divisible by 7 and 11.

⇒ (x²− 1) is divisible by 77 [77 = 7 × 11]

Multiples of 77 are 77, 154, 231, 308, 385, 462, 539, 616, 693, 770, 847, 924, 1001, 1078, 1155, 1232 …

Consider the numbers 78, 155, 232, 309, 386, 463, 540, 617, 694, 771, 848, 925, 1002, 1079, 1156, 1233 …

Among these numbers, it can be observed that 1156 (= 34²) is the first perfect square.

Hence, 1156 is the required least perfect square.

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(ANS given :235)

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