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12 Dec 2024, 14:11
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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(N/A)
Question Stats:
50% (01:13) correct
50% (01:01) wrong based on 2 sessions
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If \(A_{n} = 7^n - 1\), what is the units digit of \(A^{33}?\)
31 Dec 2024, 02:03
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\(A_{n} = 7^n − 1\)
To find \(A_{33}\) we will replace n with 33 in \(A_{n} = 7^n − 1\)
=> \(A_{33} = 7^{33} − 1\)
[ Watch this video to MASTER FUNCTIONS ]
To find the units' digit of \(7^{33}\), we need to find the cycle of units' digit of power of 7
\(7^1\) units’ digit is 7
\(7^2\) units’ digit is 9
\(7^3\) units’ digit is 3
\(7^4\) units’ digit is 1
\(7^5\) units’ digit is 7
=> The power repeats after every \(4^{th}\) power
=> Cycle of units' digit of power of 7 = 4
=> We need to divide the power by 4 and check the remainder
33 when divided by 4 gives 1 remainder
=> Units' digit of \(7^{33}\) = Units' digit of \(7^1\) = 7
=> Units' digit of \(7^{33} − 1\) = 7 - 1 = 6
So, Answer will be 6
Hope it helps!
Watch the following video to MASTER Units Digit
To find \(A_{33}\) we will replace n with 33 in \(A_{n} = 7^n − 1\)
=> \(A_{33} = 7^{33} − 1\)
[ Watch this video to MASTER FUNCTIONS ]
To find the units' digit of \(7^{33}\), we need to find the cycle of units' digit of power of 7
\(7^1\) units’ digit is 7
\(7^2\) units’ digit is 9
\(7^3\) units’ digit is 3
\(7^4\) units’ digit is 1
\(7^5\) units’ digit is 7
=> The power repeats after every \(4^{th}\) power
=> Cycle of units' digit of power of 7 = 4
=> We need to divide the power by 4 and check the remainder
33 when divided by 4 gives 1 remainder
=> Units' digit of \(7^{33}\) = Units' digit of \(7^1\) = 7
=> Units' digit of \(7^{33} − 1\) = 7 - 1 = 6
So, Answer will be 6
Hope it helps!
Watch the following video to MASTER Units Digit
31 May 2025, 03:31
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OE
The units digits of the powers of 7 follow a repeating pattern: 7, 49, 343, 2,401, etc. Pattern = {7, 9, 3, 1}. Every 4th term ends the pattern. So, every term that is a multiple of 4 will end the pattern: 4th, 8th, 12th, continuing up to the 32nd term. The pattern begins again for the 33rd term, so \(A_{33}\) has the same units digit as \(A_{1}\) , which is 7. The units digit of \(7^{33}\) is 7, and 7 − 1 = 6.
The units digits of the powers of 7 follow a repeating pattern: 7, 49, 343, 2,401, etc. Pattern = {7, 9, 3, 1}. Every 4th term ends the pattern. So, every term that is a multiple of 4 will end the pattern: 4th, 8th, 12th, continuing up to the 32nd term. The pattern begins again for the 33rd term, so \(A_{33}\) has the same units digit as \(A_{1}\) , which is 7. The units digit of \(7^{33}\) is 7, and 7 − 1 = 6.
