Find the remainder when the number represented by 22334 raised to the power $(1^{2} + 2^{2} + . . . + 66^{2})$ is divided by 5?

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Solution

The correct option is B 4
The given expression can be seen as $\frac{(22334^{O D D P O W E R})}{5}$ since the sum of $1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . + 66^{2}$ can be seen to be an odd number. The remainder would always be 4 in such a case.

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The remainder obtained when $(1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + \dots 100^{2})$ is divided by 7 is:

Find the least number which when divided by 12, leaves a remainder of 7, when divided by 15, leaves a remainder of 10 and when divided by 16, leaves a remainder of 11.

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Find the remainder when the number represented by 22334 raised to the power (12+22+...+662) is divided by 5?

The correct option is B 4The given expression can be seen as (22334ODDPOWER)5 since the sum of 12+22+32+42+...+662 can be seen to be an odd number. The remainder would always be 4 in such a case.

Find the remainder when the number represented by 22334 raised to the power $(1^{2} + 2^{2} + . . . + 66^{2})$ is divided by 5?

The correct option is B 4
The given expression can be seen as $\frac{(22334^{O D D P O W E R})}{5}$ since the sum of $1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . + 66^{2}$ can be seen to be an odd number. The remainder would always be 4 in such a case.