Author:
Least Common Multiple
Group Members:
1. (a) Find the first eight positive multiples of 315.
315, 630, 945, 1260, 1575, 1890, 2205, 2520
(b) Find the first eight positive multiples of 189.
189, 378, 567, 756, 945, 1134, 1323, 1512
(c) Find the smallest positive number that is a multiple of both 315 and 189.
945
2. (a) Find the prime factorization of 315.
315 = 3 · 105 = 3 · 5 · 21 = 3 · 5 · 3 · 7 = 32 · 5 · 7
(b) Without doing any computations determine whether or not 32 · 5 · 7 · 8 is a multiple of 315. Yes.
(c) Without doing any computations determine whether or not 32 · 5 · 7 · 19 is a multiple of 315. Yes.
(d) Without doing any computations determine whether or not 2 · 32 · 5 · 7 · 19 is a multiple of 315. Yes.
(e) Without doing any computations determine whether or not 32 · 52 · 73 is a multiple of 315. Yes.
(f) Without doing any computations determine whether or not 3 · 5 · 7 · 112 is a multiple of 315. No.
(g) Without doing any computations determine whether or not 32 · 72 is a multiple of 315. No.
(h) Explain how you determined your answers to the above questions. You should not need to explain each
one separately; the explanation you put here should work for every problem above.
We just need to make sure that the prime factorization of 315 shows up in each of the numbers. If there
are not at least two 3’s, one 5 and one 7, then the number is not a multiple of 315.
3. Find the prime factorization of 1800.
1800 = 6 · 300 = 2 · 3 · 3 · 100 = 2 · 3 · 3 · 2 · 2 · 5 · 5 = 23 · 32 · 52
4. Find the prime factorization of 2646.
2646 = 2 · 1323 = 2 · 3 · 441 = 2 · 3 · 3 · 147 = 2 · 3 · 3 · 3 · 49 = 2 · 3 · 3 · 3 · 72 = 2 · 33 · 72
5. Match each of the following numbers with the category it belongs in.
22 · 33 · 52 · 72 = 132300
23 · 34 · 52 · 72 · 11 = 8731800
This number is a multiple of 2646 but not a multiple of 1800
This number is a multiple of both 1800 and 2646
24 · 33 · 52 · 7 = 75600
This number is a multiple of 1800 but not a multiple of 2646
2 · 32 · 52 · 72 = 22050
This number is not a multiple of 1800 nor 2646
6. Suppose I have a number that is a multiple of 1800 and 2646.
(a) Could my number have exactly one 2 in its prime factorization?
there must be at least three 2’s.
No, because to be a multiple of 1800
(b) Could my number have exactly two 2’s in its prime factorization?
1800 there must be at least three 2’s.
No, because to be a multiple of
(c) Could my number have exactly three 2’s in its prime factorization?
Yes.
(d) Could my number have exactly four 2’s in its prime factorization?
Yes.
(e) Could my number have more than four 2’s in its prime factorization?
(f) Could my number have exactly one 11 in its prime factorization?
Yes.
Yes.
(g) Could my number have more than one 11 in its prime factorization?
Yes.
(h) What is the smallest amount of 3’s that my number could have in its prime factorization? To be a
multiple of 1800 I have to have at least two 3’s and to be a multiple of 2646 I have to have at least three
3’s. Therefore to satisfy both of those with the fewest amount of threes I would have to have three 3’s.
(i) What is the smallest amount of 5’s that my number could have in its prime factorization? To be a
multiple of 1800 I have to have at least two 5’s and to be a multiple of 2646 I don’t have to have any
5’s. Therefore to satisfy both of those with the fewest amount of fives I would have to have two 5’s.
(j) What is the smallest amount of 7’s that my number could have in its prime factorization? To be a
multiple of 1800 I don’t have to have any 7’s and to be a multiple of 2646 I have to have at least two
7’s. Therefore to satisfy both of those with the fewest amount of sevens I would have to have two 7’s.
(k) What is the smallest amount of 11’s that my number could have in its prime factorization? I don’t
have to have any 11’s to be a multiple of 1800 and 2646, so the least amount of 11’s would be zero.
(l) Given any other prime (i.e. other than 2,3,5,7,11), what is the smallest amount of times that prime
could show up in the prime factorization of my number? I don’t have to have any other primes to be
a multiple of 1800 and 2646, so the least amount of any other primes would be zero.
(m) If I told you I have the smallest number that is a multiple of 1800 and 2646, what’s my number?
By above the I need three 2’s, three 3’s, two 5’s and two 7’s.
23 · 33 · 52 · 72 = 264600
7. Use the prime factorization method to find the least common multiple of the following numbers.
(a) 1274 and 16660
1274 = 2 · 637 = 2 · 7 · 91 = 2 · 7 · 7 · 13 = 2 · 72 · 13
16660 = 4 · 4165 = 2 · 2 · 5 · 833 = 2 · 2 · 5 · 7 · 119 = 2 · 2 · 5 · 7 · 7 · 17 = 22 · 5 · 72 · 17
LCM = 22 · 51 · 72 · 131 · 171 = 216580
(b) 7605 and 35625
7605 = 5 · 1521 = 5 · 3 · 507 = 5 · 3 · 3 · 169 = 5 · 3 · 3 · 132 = 32 · 5 · 132
35625 = 5 · 7125 = 5 · 5 · 1425 = 5 · 5 · 5 · 285 = 53 · 5 · 57 = 54 · 3 · 19 = 3 · 54 · 19
LCM = 32 · 54 · 132 · 19 = 18061875