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C
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Difficulty:
55%
(hard)
Question Stats:
60% (01:46) correct
40%
(01:58)
wrong
based on 303
sessions
History
Date
Time
Result
Not Attempted Yet
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
Source: Self-made.
A. 61
B. 62
C. 63
D. 64
E. 65
Source: Self-made.
30 Sep 2017, 05:39
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RMD007
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
30 Sep 2017, 06:25
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RMD007
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: \(\frac{Last-First}{Increment}+ 1\) =
\(\frac{995-3}{16} + 1\) =
\(62 + 1 = 63\)
ANSWER C
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
