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How many integers from 0 to 1000 inclusive, have a remainder of 3 when  [#permalink]

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Question Stats: 66% (01:41) correct 34% (01:47) wrong based on 93 sessions

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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

Originally posted by RMD007 on 30 Sep 2017, 05:20.
Last edited by Bunuel on 01 Oct 2017, 04:12, edited 1 time in total.
Renamed the topic.
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Re: How many integers from 0 to 1000 inclusive, have a remainder of 3 when  [#permalink]

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RMD007 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

First no that will leave a remainder of 3 when divide by 16, $$a=3$$

Last no that will leave the remainder 3 when divide by 16 $$T_n= 995$$

every no from 3 to 995 will have a difference between them $$d=16$$. Hence using AP formula

$$T_n=a+(n-1)d$$ or $$995=3+(n-1)*16$$

Thus $$n=63$$

Option C
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Re: How many integers from 0 to 1000 inclusive, have a remainder of 3 when  [#permalink]

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RMD007 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

Combined approach: using arithmetic, find first and last term with remainder 3. Then calculate the number of multiples of 16 in an evenly spaced sequence.

First term: $$\frac{0}{16}$$ = 0 + R3, first term is 3

Last term: is 16(n) + 3 $$\leq{1,000}$$. Ignore the 3 for a bit.

$$\frac{1,000}{16} = 62.x$$

$$(16)(62) = 992$$. Now add 3.
$$(992+3) = 995$$ = Last Term

Number of terms: $$\frac{Last-First}{Increment}+ 1$$ =

$$\frac{995-3}{16} + 1$$ =
$$62 + 1 = 63$$

Shortcut: Above calculation shows 62 multiples of 16 in 1,000. Highest multiple + 3 is not > 1,000. Add one more multiple (because for $$\frac{1,000}{16}$$, first multiple of 16 is 16): 0 is a multiple of every integer. 62 + 1 = 63 terms
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GMAT 1: 730 Q49 V41 Re: How many integers from 0 to 1000 inclusive, have a remainder of 3 when  [#permalink]

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I did it in a more ''muscle'' way:

For the remainder to be 3 when divided by 16, must follow the equation:
3+16*n (n is integer)
n=0 -> 3
n=1 -> 19

Whats in biggest n, that the result of the equation is smaller than 1.000 (inclusive)?
Let the +3 aside for a bit..

16*10 = 160
16*20 = 320
16*40 = 640
16*60 = 640+320 = 960
16*61 = 976
16*62 = 992 (enough) + 3 = 995 (last number that follows the equation)

Answer: 62 + (n = 0) option = 63 possibilities

What you guys think?
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Re: How many integers from 0 to 1000 inclusive, have a remainder of 3 when  [#permalink]

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RMD007 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

a = 3
an = 995

a + (n-1)d = an

3 + (n-1)x16 = 995

n = 63
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Re: How many integers from 0 to 1000 inclusive, have a remainder of 3 when  [#permalink]

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RMD007 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

$$x = 16q + 3$$

$$0 \leq 16q + 3 \leq 1000$$

$$-3 \leq 16q \leq 997$$

$$-0.something \leq q \leq 62.something$$

q can take 63 integer values from 0 to 62 inclusive, thus x can also take 63 values.

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How many integers from 0 to 1000 inclusive, have a remainder of 3 when  [#permalink]

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vitorcbarbieri wrote:
I did it in a more ''muscle'' way:

For the remainder to be 3 when divided by 16, must follow the equation:
3+16*n (n is integer)
n=0 -> 3
n=1 -> 19

Whats in biggest n, that the result of the equation is smaller than 1.000 (inclusive)?
Let the +3 aside for a bit..

16*10 = 160
16*20 = 320
16*40 = 640
16*60 = 640+320 = 960
16*61 = 976
16*62 = 992 (enough) + 3 = 995 (last number that follows the equation)

Answer: 62 + (n = 0) option = 63 possibilities

What you guys think?

vitorcbarbieri

I think you found an approach with a smart pattern that works! It is a bit similar to mine, above.

I do not think your approach is "brute force" except maybe in one place where, had the upper limit of the range been an ugly number, you might have been calculating for awhile.

I didn't use round multiples of 10, times 16, to "multiply up" to the highest term. Maybe that is faster. It is certainly smart. Next time, if numbers aren't bad, I'll try your way. Just one tiny cautionary thought . . .

If you were to get some heinous number, say 173,619, as the last of the range, you might be multiplying for awhile, even if you used thousands, hundreds, and then tens. In a case such as that, to find multiples, you may want to try the reverse: divide the heinous number by the multiple to get the greatest number that "n" in 16n can be.

In other words: to find the highest multiple of 16 close to but less than 1,000: divide 1000 by 16 to get 62.xx.

It's 62.5, but I didn't calculate that. The second I saw I had a decimal coming, I just noted the .xx. Then I took the integer, 62, and multiplied by 16 to give me that highest value, 992. Added 3. Done on that front (very similar to your conclusion). If this cautionary thought seems dumb, of course, ignore it.

In any event, nice work! Glad you posted. Kudos.
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Re: How many integers from 0 to 1000 inclusive, have a remainder of 3 when  [#permalink]

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SaGa wrote:
RMD007 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

a = 3
an = 995

a + (n-1)d = an

3 + (n-1)x16 = 995

n = 63

This approach is understandable,,,, thanks,,,
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Re: How many integers from 0 to 1000 inclusive, have a remainder of 3 when  [#permalink]

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RMD007 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

The first number from 0 to 1000 inclusive that has a remainder of 3 when divided by 16 is 3, and the last number is 995.

Thus, there are (995 - 3)/16 + 1 = 63 integers from 0 to 1000 inclusive that have a remainder of 3 when divided by 16.

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Re: How many integers from 0 to 1000 inclusive, have a remainder of 3 when  [#permalink]

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