Divisibility and Primes
44. abc is a three-digit number in which a is the hundreds digit, b is the tens digit, and c is the units digit. Let &(abc)& = 2^a * 3^b *5^c. For example, &(203)& = 2^2*3^0*5^3 = 500. For how many three-digit numbers abc will the function &(abc)& yield a prime number?
Answer: 1 2 ok, 100 ok 3 ok, 010 no 5 ok, 001 no
Quantity A = The number of distinct prime factors of 100,000 Quantity B = The number of distinct prime factors of 99,000
2, 5 3,11,2,5 B
! If n is an integer and n3 is divisible by 24, what is the largest number that must be a factor of n?
Because n is an integer, for every prime factor n has, n^3 must have three of them. n^3 is divisible by 24, so n^3 is div by 2*2*2*3 Since n^3 is a cube, it is div by 3 3s and 3 2s. n must be div by 2 × 3 = 6.
числитель знаменатель частное
quotient=numerator/denominator
For which two of the following values is the product a multiple of 27? Indicate two such values. 1 7 20 28 63 217 600 700
27=3*3*3 63=9*3 600=3*200
If 3^x*5^2 is divided by 3^5*5^3, the quotient terminates with one decimal digit. If x > 0, which of the following statements must be true? (A) x is even (B) x is odd (C) x < 5 (D) x >= 5 (E) x = 5
D When a non-multiple of 3 is divided by 3, the quotient does not terminate (for instance, 1/3 = 0.333... x must be large enough to cancel out 3^5 in the denominator x must be at least 5 E works as well, D must be true
What is the remainder when 13^17 + 17^13 is divided by 10?
Ignore all but the units digits What is the units digit of 3^17 + 7^13? 3 9 7 1 => 17th term ends in 3 7 9 3 1 => 13th term ends in 7 ...3+...7=...10:10=0 Answer: 0
! a, b, c, and d are positive integers. If a/b has a remainder of 9 and c/d has a remainder of 10, what is the minimum possible value for bd?
When dividing, the remainder is always less than the divisor. If you divided a by b to get a remainder of 9, then b must have been greater than 9. Similarly, d must be greater than 10. Since b and d are integers, the smallest they could be is 10 and 11, respectively. Thus, the minimum that bd could be is 10 × 11 = 110. As an example, try a = 19, b = 10, c = 21, and d = 11 (generate a by adding remainder 9 to the value of b, and generatec by adding remainder 10 to the value of d.) It is not possible to generate an example in which any of the four numbers are smaller. The least possible value of bd is 110.
