(a) how many integers in the range 1 through 120 are integer multiples of 2, 3, or 5? keyboard_arrow_down solution (b) how many integers in the range 1 through 140 are integer multiples of 2, 5, or 7?

(a) 88 integers

(b) 92 integers

Step-by-step explanation:

(a) integers whose last digits are divisible by 2 are multiples of two or numbers whose digits ends with zero. So for number 1-120 , all the even numbers which are sixty in number are are multiples of two.

For  3, numbers whose digits sum is divisible by three are multiples of three. 3,6,9,12,15,18,21,24,27,30 are multiples of three from numbers 1-30. we have four 30s in 120. which means numbers of integers will be 10*4 = 40integers. However out of these numbers , half are also integers of 2 which reduces the number added to 20integers.

For 5, numbers whose digits ends with 5 or 0 are multiples of 5. this gives us 24 integers for 1-120. but out of these 24integers, 16 are common integers of 2 and 3 which reduces the number added to 8integers.

Thus from 1-120 the intergers of 2,3 or 5 = 60+20+8 = 88integers.

(b) if we are considering from numbers 1-140;

for 2 we wil have 70 integers,

for 5 we will have 28 integers, but those integers that end with 0 are also integers of 2 which reduces the number added to 14.

For 7, numbers 7,14,21,28,35,42,49,56,63,70 are multiples of three from 1-70. This pattern is repeated from number 71-140. hence we have 20 integers in all. However 12 of the multiples are also multiples of either 2 or 5 which reduces the number to 8 integers.

Thus from 1-140, the integers of 2, 5, or 7 = 70+14+8 = 92integers

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)