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09 Jun 2020, 07:13
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Difficulty:
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A function D(a, 10b + c) is defined as the remainder when the sum \(a^0 + a^1 + ... + a^{10b + c}\) is divided by c, where a, b and c are single-digit positive integers. What is the value of D(y, 10x + z) where x, y and z are single-digit positive integers such that x < y < z , x and z are perfect squares and the difference between the sum and the product of the prime factors of y is 1?
A. 0
B. 3
C. 6
D. 7
E. Cannot be determined
Are You Up For the Challenge: 700 Level Questions
A. 0
B. 3
C. 6
D. 7
E. Cannot be determined
Are You Up For the Challenge: 700 Level Questions
09 Jun 2020, 07:58
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y must be equal to 6. [(3*2)-(3+2) =1]
since z is a perfect square and greater than y=6, z must be equal to 9.
x can be 1 or 4
Case 1- x= 1
\(6^n\) must be divisible by 9, for n>1
hence, you can write
\([6^0+6^1+6^2+........+6^{19}]\) = 6^0 + 6 + 9k = 9k+7
D(6, 19) = 7
Case 2- x= 4
Similarly we can write
\([6^0+6^1+6^2+........+6^{49}]\) = 1+6+9m = 9m+7
D(6, 49) = 7
Answer - D
since z is a perfect square and greater than y=6, z must be equal to 9.
x can be 1 or 4
Case 1- x= 1
\(6^n\) must be divisible by 9, for n>1
hence, you can write
\([6^0+6^1+6^2+........+6^{19}]\) = 6^0 + 6 + 9k = 9k+7
D(6, 19) = 7
Case 2- x= 4
Similarly we can write
\([6^0+6^1+6^2+........+6^{49}]\) = 1+6+9m = 9m+7
D(6, 49) = 7
Answer - D
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09 Jun 2020, 08:30
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[quote="nick1816"]y must be equal to 6. [(3*2)-(3+2) =1]
since z is a perfect square and greater than y=6, z must be equal to 9.
x can be 1 or 4
Case 1- x= 1
\([6^0+6^1+6^2+........+6^{19}]\) mod 9 = \([6^0+6]\) mod 9 = 7 mod 9
Case 2- x= 4
\([6^0+6^1+6^2+........+6^{49}]\) mod 9 = \([6^0+6]\) mod 9 = 7 mod 9
D
would you please elaborate on the procedure of finding mod?
since z is a perfect square and greater than y=6, z must be equal to 9.
x can be 1 or 4
Case 1- x= 1
\([6^0+6^1+6^2+........+6^{19}]\) mod 9 = \([6^0+6]\) mod 9 = 7 mod 9
Case 2- x= 4
\([6^0+6^1+6^2+........+6^{49}]\) mod 9 = \([6^0+6]\) mod 9 = 7 mod 9
D
would you please elaborate on the procedure of finding mod?
