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GCD & LCM: all problems

Focus mode: every practice problem from the chapter, one after another.

0 / 17 solved

Problem 23.1.1

AMC 10

Let a,b,ca, b, c, and dd be positive integers such that gcd⁡(a,b)=24\gcd(a, b) = 24, gcd⁡(b,c)=36\gcd(b, c) = 36, gcd⁡(c,d)=54\gcd(c, d) = 54, and 70<gcd⁡(d,a)<10070 < \gcd(d, a) < 100. Which of the following must be a divisor of aa?

Click a choice or type its letter (A–E).

Problem 23.1.2

IMO

This problem also appears in the book as 22.1.4.

Prove that the fraction 21n+414n+3\dfrac{21n+4}{14n+3} is irreducible for every natural number nn.

gcd(21n + 4, 14n + 3) divides 1; at n = 1 the steps are 25 and 17. Steps 25 and 17: x·25 + y·17 reaches exactly the multiples of gcd = 1

n1
target—
−60−40−2002040600
0·25 + 0·17 = 0

At n = 1: a = 25, b = 17.

Euclid on the expressions
gcd(21n + 4, 14n + 3) = ? (hidden until the end of the example)
21n + 4 = 25, 14n + 3 = 17

The token is at 0·25 + 0·17 = 0.

This one asks for a proof. Write yours on paper, use the hints if you need them, then compare it with the written proof.

Problem 23.1.3

AIME

This problem also appears in the book as 22.1.5.

What is the largest positive integer nn for which n3+100n^{3} + 100 is divisible by n+10n + 10?

Problem 23.1.4

AMC 10

How many ordered pairs (a,b)(a, b) of positive integers satisfy the equation

a⋅b+63=20⋅lcm⁡(a,b)+12⋅gcd⁡(a,b),a \cdot b + 63 = 20 \cdot \operatorname{lcm}(a, b) + 12 \cdot \gcd(a, b),

where gcd⁡(a,b)\gcd(a, b) denotes the greatest common divisor of aa and bb, and lcm⁡(a,b)\operatorname{lcm}(a, b) denotes their least common multiple?

Problem 23.1.5

AMC 10/12

Suppose aa, bb, and cc are positive integers such that

a14+b15=c210.\frac{a}{14} + \frac{b}{15} = \frac{c}{210}.

Which of the following statements are necessarily true?

I. If gcd⁡(a,14)=1\gcd(a, 14) = 1 or gcd⁡(b,15)=1\gcd(b, 15) = 1 or both, then gcd⁡(c,210)=1\gcd(c, 210) = 1.

II. If gcd⁡(c,210)=1\gcd(c, 210) = 1, then gcd⁡(a,14)=1\gcd(a, 14) = 1 or gcd⁡(b,15)=1\gcd(b, 15) = 1 or both.

III. gcd⁡(c,210)=1\gcd(c, 210) = 1 if and only if gcd⁡(a,14)=gcd⁡(b,15)=1\gcd(a, 14) = \gcd(b, 15) = 1.

Click a choice or type its letter (A–E).

Problem 23.1.6

AMC 12

Suppose that a,b,ca, b, c, and dd are positive integers satisfying all of the following relations:

abcd=26⋅39⋅57lcm⁡(a,b)=23⋅32⋅53lcm⁡(a,c)=23⋅33⋅53lcm⁡(a,d)=23⋅33⋅53lcm⁡(b,c)=21⋅33⋅52lcm⁡(b,d)=22⋅33⋅52lcm⁡(c,d)=22⋅33⋅52\begin{aligned} abcd &= 2^{6} \cdot 3^{9} \cdot 5^{7} \\ \operatorname{lcm}(a, b) &= 2^{3} \cdot 3^{2} \cdot 5^{3} \\ \operatorname{lcm}(a, c) &= 2^{3} \cdot 3^{3} \cdot 5^{3} \\ \operatorname{lcm}(a, d) &= 2^{3} \cdot 3^{3} \cdot 5^{3} \\ \operatorname{lcm}(b, c) &= 2^{1} \cdot 3^{3} \cdot 5^{2} \\ \operatorname{lcm}(b, d) &= 2^{2} \cdot 3^{3} \cdot 5^{2} \\ \operatorname{lcm}(c, d) &= 2^{2} \cdot 3^{3} \cdot 5^{2} \end{aligned}

What is gcd⁡(a,b,c,d)\gcd(a, b, c, d)?

Click a choice or type its letter (A–E).

Problem 23.1.7(additional)

AMC 12

Let NN be the second smallest positive integer that is divisible by every positive integer less than 77. What is the sum of the digits of NN?

Problem 23.1.8(additional)

AMC 10

There are 1010 horses, named Horse 11, Horse 22, …\ldots, Horse 1010. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse kk runs one lap in exactly kk minutes. At time 00 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time S>0S > 0, in minutes, at which all 1010 horses will again simultaneously be at the starting point is S=2520S = 2520. Let T>0T > 0 be the least time, in minutes, such that at least 55 of the horses are again at the starting point. What is the sum of the digits of TT?

Problem 23.1.9(additional)

AMC 12

Suppose aa, bb, cc are positive integers such that

a+b+c=23a + b + c = 23

and

gcd⁡(a,b)+gcd⁡(b,c)+gcd⁡(c,a)=9.\gcd(a, b) + \gcd(b, c) + \gcd(c, a) = 9.

What is the sum of all possible distinct values of a2+b2+c2a^{2} + b^{2} + c^{2}?

Problem 23.1.10(additional)

AMC 12

This problem also appears in the book as 23.1.7.

Let NN be the second smallest positive integer that is divisible by every positive integer less than 77. What is the sum of the digits of NN?

Problem 23.1.11(additional)

AMC 12

Let nn be the smallest positive integer such that nn is divisible by 2020, n2n^{2} is a perfect cube, and n3n^{3} is a perfect square. What is the number of digits of nn?

Problem 23.1.12(additional)

AMC 12

Mary chose an even 44-digit number nn. She wrote down all the divisors of nn in increasing order from left to right: 1,2,…,n2,n1, 2, \ldots, \dfrac{n}{2}, n. At some moment Mary wrote 323323 as a divisor of nn. What is the smallest possible value of the next divisor written to the right of 323323?

Problem 23.1.13(additional)

AMC 12

How many positive integers nn are there such that nn is a multiple of 55, and the least common multiple of 5!5! and nn equals 55 times the greatest common divisor of 10!10! and nn?

Problem 23.1.14(additional)

AIME

For how many ordered pairs of positive integers (x,y)(x, y), with y<x≤100y < x \le 100, are both xy\dfrac{x}{y} and x+1y+1\dfrac{x+1}{y+1} integers?

Problem 23.1.15(additional)

AMC 10

Let nn be the least positive integer greater than 10001000 for which

gcd⁡(63,n+120)=21andgcd⁡(n+63,120)=60.\gcd(63, n + 120) = 21 \quad\text{and}\quad \gcd(n + 63, 120) = 60.

What is the sum of the digits of nn?

Problem 23.1.16(additional)

AIMEChallenge

Consider the sequence (ak)k≥1(a_k)_{k \ge 1} of positive rational numbers defined by a1=20202021a_1 = \dfrac{2020}{2021} and for k≥1k \ge 1, if ak=mna_k = \dfrac{m}{n} for relatively prime positive integers mm and nn, then

ak+1=m+18n+19.a_{k+1} = \frac{m + 18}{n + 19}.

Determine the sum of all positive integers jj such that the rational number aja_j can be written in the form tt+1\dfrac{t}{t+1} for some positive integer tt.

Problem 23.1.17(additional)

AIMEChallenge

Find the number of ordered pairs (m,n)(m, n) such that mm and nn are positive integers in the set {1,2,…,30}\{1, 2, \ldots, 30\} and the greatest common divisor of 2m+12^{m} + 1 and 2n−12^{n} - 1 is not 11.