How many positive multiples of 13 are three-digit integers?

URGENT

here are 900 three digit numbers, from 100 to 999 (999–100+1 = 900). If we divide 900 by 13, we get 69. Since these are the multiples of 13, they are bound to be sequence of odd and even multiples. We have 68/2 = 34 odd multiples and 34 even multiples. So, the question is, whether the last multiple is odd or even. Since the first 3 digit multiple of 13 is 104, which is an even number, the last or 69 number is also going to be even number.

Thus, we have 34 + 1 = 35, three digit even numbers that are divisible by 13.

The required answer is 35.

Step-by-step explanation:

Correct me if this is wrong

The smallest 3-digit integer is 100, the largest is 999.

Divide 100 by 13; the quotient is 7 and the remainder is 9; this means that the first multiple of 13 that will be at least 100 is 8.

Divide 1000 by 13; the quotient is 76 and the remainder is 12; this means that the last multiple fo 13 that will be smaller than 1000 is 76.

We now need to count the number of numbers starting with 8 and ending with 76.

A short way to do this is to subtract 8 from 76 and add 1 -- 76 - 8 + 1 = 69. (Don't forget to add 1!)

There are 69 positive integer multiples of 13 whose answers have 3 digits.

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that .

also, it is given that

so, from the options, we need that .

a.

b.

c.  

d.

as, we can see that only option d gives the same ratio as i.e. .

hence, the possible lengths are bc = 12 and ef = 16.

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step-by-step explanation: