Problem Solving

How many integers from 0 to 1000 inclusive, have a remainder of 3 when divided by 16?

A. 61
B. 62
C. 63
D. 64
E. 65
QA is c
Can some experts help me with this? How to come up with the correct answer?
Thanks

Roland2rule wrote:How many integers from 0 to 1000 inclusive, have a remainder of 3 when divided by 16?

A. 61
B. 62
C. 63
D. 64
E. 65
QA is c
Can some experts help me with this? How to come up with the correct answer?
Thanks

Call the integers that give a remainder of 3 when divided by 16: 16x + 3

Smallest number in the range that gives a remainder of 3 when divided by 16: 3; (when x = 0)

Largest number in the range that gives a remainder of 3 when divided by 16: 995; (when x = 62) note: it's easy to see that 16*60 + 3 = 963. From there, it's straightforward to extrapolate the largest possible value. 16*61 + 3 = 979; 16*62 + 3 = 995.

Number of integers between 0 and62 inclusive: 63. The answer is C

Hello Roland2rule.

The idea here is to find a number N such that $$16\cdot N+3\le1000\ and\ N\ge0.$$

Let's try some test values:

$$N=50=>16\cdot50+3=803\ \le\ 1000.$$

This tell us that all there are 51 integers that satisfy the condition above. Now, let's try with a greater N.

$$N=60\ =>\ 16\cdot60+3=963\le1000.$$

If we try with N=62 and N=63 we will get

$$N=62\ =>\ 16\cdot62+3=995\le1000$$ $$N=63\ =>\ 16\cdot63+3=1011>1000$$

So, the 63 doesn't satisfy the condition.

In conclusion, all the integers from 0 to 62 satisfy the condition, that is to say, there are 63 integers that have a remainder of 3 when divided by 16.

The correct answer is C.

I hope this answer can help you.

Regards.