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01 Nov 2020, 00:30
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
51% (02:07) correct
49%
(02:02)
wrong
based on 227
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Not Attempted Yet
If \(Z = 1^1 + 2^2 + 3^3 +...+ 8^8\), what is the remainder when Z is divided by 10?
(A) 0
(B) 2
(C) 4
(D) 6
(E) 8
(A) 0
(B) 2
(C) 4
(D) 6
(E) 8
01 Nov 2020, 03:07
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Bunuel
The question is simply asking what is the last digit of Z.
We use cyclicity to find the last digit of each number. Cyclicity means after how many cycles the base comes back as itself.
Cyclicity of 0, 1, 5 and 6 is 1.
Cyclicity of 4 and 9 is 2.
Cyclicity of 2, 3, 7 and 8 is 4.
Last digits of \(1^1\), \(2^2\) and \(3^3\) are 1, 4 and 7
\(4^4\) can be written as \((4^2)^2 = 16^2\). Therefore the last digit is 6.
The last digit of \(5^5\) and \(6^6\) are 5 and 6 respectively as their cyclicity is 1.
\(7^7\) can be written as \(7^4 * 7^3\).
\(7^4 = 7^2 * 7^2\) = 49 * 49 and the last digit of this number is 1. We only see the last digit of \(7^3\) which is 3.
\(8^8 = (8^4)^2\).
\(8^4 = 8^2 * 8^2\) which is 64 * 64, which ends with 6 and \(6^2\) will be ending with 6.
The sum now becomes 1 + 4 + 7 + 6 + 5 + 6 + 3 + 6 = 38 and the last digit is 8 which is the remainder.
Option E
Arun Kumar
06 Nov 2022, 04:51
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Given that \(Z = 1^1 + 2^2 + 3^3 +...+ 8^8\) and we need to find the remainder when Z is divided by 10?
Theory
=> Remainder of \(Z = 1^1 + 2^2 + 3^3 +...+ 8^8\) by 10
= Remainder of \(1^1\) by 10 + Remainder of \(2^2\) by 10 +... + Remainder of \(8^8\) by 10
= Unit's Digit \(1^1\) + Unit's Digit \(2^2\) +... + Unit's Digit \(8^8\)
Unit's digit of \(1^1\) = 1
Unit's digit of \(2^2\) = 4
Unit's digit of \(3^3\) = 7
Unit's digit of \(4^4\) = 6
Unit's digit of \(5^5\) = 5 (Unit's digit of any positive integer power of 5 = 5)
Unit's digit of \(6^6\) = 6 (Unit's digit of any positive integer power of 6 = 6)
Unit's digit of \(7^7\)
\(7^1\) units’ digit is 7 [ 7 ]
\(7^2\) units’ digit is 9 [ 7*7 = 49 ]
\(7^3\) units’ digit is 3 [ 9*7 = 63 ]
\(7^4\) units’ digit is 1 [ 3*7 = 21 ]
\(7^5\) units’ digit is 3 [ 1*7 = 7 ]
So, unit's digit of power of 7 repeats after 4
=> Unit's digit of \(7^7\) = \(7^3\) = 3
Unit's digit of \(8^8\)
\(8^1\) units’ digit is 8 [ 8 ]
\(8^2\) units’ digit is 4 [ 8*8 = 64 ]
\(8^3\) units’ digit is 2 [ 4*8 = 32 ]
\(8^4\) units’ digit is 6 [ 2*8 = 16 ]
\(8^5\) units’ digit is 8 [ 6*8 = 8 ]
So, unit's digit of power of 8 repeats after 4
=> Unit's digit of \(8^8\) = \(8^4\) = 6
=> Unit's Digit \(1^1\) + Unit's Digit \(2^2\) +... + Unit's Digit \(8^8\) = Unit's digit of (1 + 4 + 7 + 6 + 5 + 6 + 3 + 6)
= Unit's digit of 38 = 8
So, Answer will be E
Hope it helps!
Watch the following video to MASTER Remainders
Theory
- - Remainder of a number by 10 is same as the unit's digit of the number
- Remainder of sum of two numbers = sum of their remainders
=> Remainder of \(Z = 1^1 + 2^2 + 3^3 +...+ 8^8\) by 10
= Remainder of \(1^1\) by 10 + Remainder of \(2^2\) by 10 +... + Remainder of \(8^8\) by 10
= Unit's Digit \(1^1\) + Unit's Digit \(2^2\) +... + Unit's Digit \(8^8\)
Unit's digit of \(1^1\) = 1
Unit's digit of \(2^2\) = 4
Unit's digit of \(3^3\) = 7
Unit's digit of \(4^4\) = 6
Unit's digit of \(5^5\) = 5 (Unit's digit of any positive integer power of 5 = 5)
Unit's digit of \(6^6\) = 6 (Unit's digit of any positive integer power of 6 = 6)
Unit's digit of \(7^7\)
\(7^1\) units’ digit is 7 [ 7 ]
\(7^2\) units’ digit is 9 [ 7*7 = 49 ]
\(7^3\) units’ digit is 3 [ 9*7 = 63 ]
\(7^4\) units’ digit is 1 [ 3*7 = 21 ]
\(7^5\) units’ digit is 3 [ 1*7 = 7 ]
So, unit's digit of power of 7 repeats after 4
=> Unit's digit of \(7^7\) = \(7^3\) = 3
Unit's digit of \(8^8\)
\(8^1\) units’ digit is 8 [ 8 ]
\(8^2\) units’ digit is 4 [ 8*8 = 64 ]
\(8^3\) units’ digit is 2 [ 4*8 = 32 ]
\(8^4\) units’ digit is 6 [ 2*8 = 16 ]
\(8^5\) units’ digit is 8 [ 6*8 = 8 ]
So, unit's digit of power of 8 repeats after 4
=> Unit's digit of \(8^8\) = \(8^4\) = 6
=> Unit's Digit \(1^1\) + Unit's Digit \(2^2\) +... + Unit's Digit \(8^8\) = Unit's digit of (1 + 4 + 7 + 6 + 5 + 6 + 3 + 6)
= Unit's digit of 38 = 8
So, Answer will be E
Hope it helps!
Watch the following video to MASTER Remainders
