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What is remainder when 5 to the power of 30 is divided by 7?
Look at the powers of 5 mod 7: 5¹ mod 7 = 5 5² mod 7 = 5 × (5¹ mod 7) mod 7 = (5 × 5) mode 7 = 25 mod 7 = 4 5³ mod 7 = 5 × (5² mod 7) mod 7 = (5 × 4) mod 7 = 20 mod 7 = 6 5⁴ mod 7 = 5 × (5³ mod 7) mod 7 = (5 × 6) mod 7 = 30 mod 7 = 2 5⁵ mod 7 = 5 × (5⁴ mod 7) mod 7 = (5 × 2) mod 7 = 10 mod 7 = 3 5⁶ mod 7 = 5 × (5⁵ mod 7) mod 7 = (5 × 3) mod 7 = 15 mod 7 = 1 5⁷ mod 7 = 5 × (5⁶ mod 7) mod 7 = (5 × 1) mod 7 = 5 mod 7 = 5 At this point, it is obvious that the remainders will repeat the cycle {5, 4, 6, 2, 3, 1} There are 6 remainders in the cycle, so the remainder of 30 divided by 6 will tell you which remainder to use; if the remainder is 0, use the 6th element. 30 ÷ 6 = 5 r 0 →use the 6th element which is 1, so 5³⁰ ÷ 7 will have a remainder of 1. 1 ≡ 5³⁰ mod 7.
Why does -24 mod 7 equal 4 but 24 mod 7 equal 3?
The modulus of a number relative to a base is the positive remainder when itis divided by that base. So -24 mod 7 = (-28 + 4) mod 7 = -28 mod 7 + 4 mod 7 = 4 mod 7 = 4while 24 mod 7 = (21 + 3) mod 7 = 21 mod 7 + 3 mod 7 = 3 mod 7 = 3.
What is the value mod 101 of 1964?
1964 mod 101 = 45
Can you find a square number that when divided by 3 has a remainder of 2?
No because it is impossible. Let mod(x.3) denote the remainder when x is divided by 3. Let n be any integer. Then mod(n,3) = 0,1 or 2. When mod(n,3) = 0, mod(n2,3) = 0 When mod(n,3) = 1, mod(n2,3) = 1 When mod(n,3) = 2, mod(n2,3) = 4 and, equivalently mod(n2,3) = 1. So, there are no integers whose squar leaves a remainder of 2 when divided by 3.
What is the unit digit in 3 power 99?
To find the units digit of 399 the question being asked is: What is (399) MOD 10? This does not necessitate evaluation of 399 before the modulus is done, as it can be done whenever it is possible during the multiplication as any multiple of 10 multiplied by 3 is still a multiple of 10. The first few powers of 3 modulus 10 are: 31 MOD 10 = 3 32 MOD 10 = (3 x 31) MOD 10 = (3 x 3) MOD 10 = 9 33 MOD 10 = (3 x 32) MOD 10 = (3 x 9) MOD 10 = 27 MOD 10 = 7 34 MOD 10 = (3 x 33) MOD 10 = (3 x 7) MOD 10 = 81 MOD 10 = 1 35 MOD 10 = (3 x 34) MOD 10 = (3 x 1) MOD 10 = 3 36 MOD 10 = (3 x 35) MOD 10 = (3 x 3) MOD 10 = 9 At this point, it can be seen that the answer is a repeating pattern of 3, 9, 7, 1, 3, 9, ... So we need the 99th element of this pattern. The pattern is a repeat of 4 digits, so we calculate 99 MOD 4 = 3. So the 3rd element of the repeating part is the answer: 7. (If the power MOD 4 had been 0, it would have been the 4th element of the pattern: 1)
