The digit in the unit's place of the number 17^1999 + 11^1999 – 7^1999 is
Units digits of powers of 7 repeat in the cycle 7, 9, 3, 1. Since 1999 leaves remainder 3 when divided by 4, both 17^1999 and 7^1999 end in 3, and any power of 11 ends in 1.…
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The digit in the unit's place of the number 17^1999 + 11^1999 – 7^1999 is
Answer and explanation
Correct answer D
Units digits of powers of 7 repeat in the cycle 7, 9, 3, 1. Since 1999 leaves remainder 3 when divided by 4, both 17^1999 and 7^1999 end in 3, and any power of 11 ends in 1.
The units digit of the expression is 3 + 1 - 3 = 1. As 17^1999 is larger than 7^1999, no borrowing arises, so the digit is 1, which is none of 3, 7 or 9.
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Uttarakhand PCS (Combined State Civil Services)