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29 Jan 2025, 01:30
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C
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Difficulty:
15%
(low)
Question Stats:
82% (01:19) correct
18%
(01:35)
wrong
based on 213
sessions
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What are the last two digits (at ten's place and unit's place) in the numerical value of \(7^{507}\)?
A. 01
B. 07
C. 43
D. 49
E. 81
A. 01
B. 07
C. 43
D. 49
E. 81
29 Jan 2025, 03:32
Kudos
Bookmarks
Well luckily this can be checked through cyclicity of 7.
The pattern of units digit is :
\(7^1=7\)
\(7^2=_9\)
\(7^3=_3\)
\(7^4=_1\)
\(7^5=_7\)
Hence the unit digit cycles after every 4 powers of 7.
Now checking,
507 = 4*126 (=504) + 3
Hence the units digit will be the 3rd digit in the cycle i.e _3.
The only option with units digit with 3 is 43.
(If the power were some power with units digit were 1, and there are 2 options with units digit 1. Then we would have to track both tens and units digits both, Both these have cyclicity of 4 powers. Hence I started with we are Lucky)
IMO Ans C
The pattern of units digit is :
\(7^1=7\)
\(7^2=_9\)
\(7^3=_3\)
\(7^4=_1\)
\(7^5=_7\)
Hence the unit digit cycles after every 4 powers of 7.
Now checking,
507 = 4*126 (=504) + 3
Hence the units digit will be the 3rd digit in the cycle i.e _3.
The only option with units digit with 3 is 43.
(If the power were some power with units digit were 1, and there are 2 options with units digit 1. Then we would have to track both tens and units digits both, Both these have cyclicity of 4 powers. Hence I started with we are Lucky)
IMO Ans C
29 Jan 2025, 20:46
Kudos
Bookmarks
This is a cyclicity problem.
As it knows, when a number (in this case 7) is raised to the n power, the unit digit follows a pattern as per its power cycle.
7 have cyclicity of 4 because every four numbers the unit digits repeat
\(7^1 \) ends in 7 as\( 7^5 \)ends in 7 -----\(-7^(4k+1)\)
\(7^2 \)ends in 9 as \(7^6 \)ends in 7 ------\(7^(4k+2)\)
\(7^3\) ends in 3 as \(7^7 \)ends in 3 ------\(7^(4k+3)\)
\(7^4 \)ends in 1 as \(7^8 \)ends in 7 ------\(7^(4k)\)
So for find the solution, we must just express 507 in the form 4k+n , and this can be obtaining by finding the reminder of 507/4 that is 3, so we can express \(7^(507)\) as \((7)^(4k+3)\) and the last digits will be the same as \(7^3 \): 343
43 is the answer: C
As it knows, when a number (in this case 7) is raised to the n power, the unit digit follows a pattern as per its power cycle.
7 have cyclicity of 4 because every four numbers the unit digits repeat
\(7^1 \) ends in 7 as\( 7^5 \)ends in 7 -----\(-7^(4k+1)\)
\(7^2 \)ends in 9 as \(7^6 \)ends in 7 ------\(7^(4k+2)\)
\(7^3\) ends in 3 as \(7^7 \)ends in 3 ------\(7^(4k+3)\)
\(7^4 \)ends in 1 as \(7^8 \)ends in 7 ------\(7^(4k)\)
So for find the solution, we must just express 507 in the form 4k+n , and this can be obtaining by finding the reminder of 507/4 that is 3, so we can express \(7^(507)\) as \((7)^(4k+3)\) and the last digits will be the same as \(7^3 \): 343
43 is the answer: C
