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What are the last two digits of 583^512

Give the logic. not the answer from calculator.

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Give the logic. not the answer from calculator.

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Let N = (500+83)⁵¹² Using binomial expansion,

N = Sum of terms that have powers of 500 as a factor + 83⁵¹²

So on the R H S, only the last term contributes to the last two digits of N. Other terms have at least two zeros at the tenths and units positions. Last two digits of N are the last two digits of M = 83⁵¹².

M = 83⁵¹² =

Now, the last two digits of the numbers which are powers of 3 are :

83⁴: 21 83⁸: 41 83¹⁶: 81 83³² : 61

83⁶⁴ : 21 83¹²⁸ : 41 83²⁵⁶ : 81 83⁵¹² : 61

61 is last two digits of 83⁵¹² and 583⁵¹²

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(80+3)^512 : the last two terms in the binomial expansion are:

512 * 80 * 3^511 + 3^512

These have the last two digits as 21 and 41 respectively. So the total is 61.

N = Sum of terms that have powers of 500 as a factor + 83⁵¹²

So on the R H S, only the last term contributes to the last two digits of N. Other terms have at least two zeros at the tenths and units positions. Last two digits of N are the last two digits of M = 83⁵¹².

M = 83⁵¹² =

Now, the last two digits of the numbers which are powers of 3 are :

83⁴: 21 83⁸: 41 83¹⁶: 81 83³² : 61

83⁶⁴ : 21 83¹²⁸ : 41 83²⁵⁶ : 81 83⁵¹² : 61

61 is last two digits of 83⁵¹² and 583⁵¹²

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(80+3)^512 : the last two terms in the binomial expansion are:

512 * 80 * 3^511 + 3^512

These have the last two digits as 21 and 41 respectively. So the total is 61.