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08 Apr 2018, 07:00
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whenever you come across a question like this :-- a number raised to higher power of integer divided by 2,3,4,5,6,7,8,9..
and asking for remainder...then, the question is actually asking for the unit digit of the numerator..
now,
3^4 has a unit digit of 1 => unit digit of 1 will repeat after every fourth power of 3.
6 to the power anything, always has a unit digit of 6
7^4 also has a unit digit of 1 => unit digit of 1 will repeat after every fourth power of 7
now, 151 = 148 + 3
7^148 * 7^3 = unit digit of 1 * unit digit of 3 = unit digit of 3..
any number divided by 5 gives remainder = unit digit of the number, when unit digit is less than 5
and (unit digit - 5) , when unit digit is greator than 5
and asking for remainder...then, the question is actually asking for the unit digit of the numerator..
now,
3^4 has a unit digit of 1 => unit digit of 1 will repeat after every fourth power of 3.
6 to the power anything, always has a unit digit of 6
7^4 also has a unit digit of 1 => unit digit of 1 will repeat after every fourth power of 7
now, 151 = 148 + 3
7^148 * 7^3 = unit digit of 1 * unit digit of 3 = unit digit of 3..
any number divided by 5 gives remainder = unit digit of the number, when unit digit is less than 5
and (unit digit - 5) , when unit digit is greator than 5
09 Apr 2018, 16:31
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blackcrow
To determine the remainder when 7^131 is divided by 5, we need only to divide the units digit of 7^131 by 5.
Let’s determine the units digit of 7^131:
When writing out the pattern of 7 raised to positive integer exponents, notice that we are concerned ONLY with the units digit of 7 raised to each power.
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7
The pattern of the units digits of powers of 7 repeats every 4 exponents. The pattern is 7–9–3–1. In this pattern, we see that all positive exponents that are multiples of 4 will produce a 1 as its units digit. Thus:
7^132 has a units digit of 1 and so 7^131 has units digit of 3. So, the remainder of 3/5 = 3.
Answer: D
15 Aug 2019, 07:52
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blackcrow
\(rof: 7^{131}/5\)
\(rof: (5+2)^{131}/5\)
\(rof: (2)^{131}/5\)
cyclicity rof \(2^{n}/5={2/5, 4/5, 8/5, 16/5…} = {2,4,3,1…}\)
\(rof: (2)^{131}/5\) is \(131/4=3rd.position=3\)
Answer (D).
