Around the World in 80 Questions (Day 6): How many three-digit

Question

How many three-digit positive integers are multiples of 7 but not multiples of 6 or 15 ?

A. 98
B. 102
C. 106
D. 110
E. 114

Bunuel · Accepted Answer

How many three-digit positive integers are multiples of 7 but not multiples of 6 or 15?

A. 98
B. 102
C. 106
D. 110
E. 114

Let's determine the number of multiples of 7 from 100 to 1000, inclusive. The number of multiples of an integer within a range can be calculated using the following formula:

•  \frac{	ext{last multiple in the range - first multiple in the range} }{	ext{multiple} }+1

Thus:

• The number of multiples of 7 in the given range is  \frac{ last - first}{multiple}+1=\frac{994 -105}{7}+1=128

Since we need only those multiples that are not also multiples of 6 or 15, we should subtract from that number the multiples of 42, which is the least common multiple of 7 and 6, and the multiples of 105, which is the least common multiple of 7 and 15.

• The number of multiples of 42 in the given range is  \frac{ last - first }{ multiple}+1=\frac{966-126}{42}+1=21

• The number of multiples of 105 in the given range is  \frac{ last - first }{ multiple}+1=\frac{945-105}{42}+1=9

However,  both  counts above include the multiples of 6, 7, and 15, namely, the multiples of 210. Hence, we need to subtract that number from 21 + 9 = 30 to avoid double-counting those:

• The number of multiples of 210 in the given range is  \frac{ last - first}{multiple}+1=\frac{840 -210}{210}+1=4

Therefore, the number of three-digit positive integers that are multiples of 7 but not multiples of 6 or 15 is  128 - (21 + 9 - 4) = 102 .

Answer: B

vedha0 · Answer

answer is B

first 3 digit multiple of 7 = 105
last 3 digit multiple of 7 = 994
No of multiples of 7 = ((994-105)/7) + 1 = 127 + 1 = 128

we need to find those multiples of 7 which are NOT multiples of 6 and 15. so subtract multiples of 42 and 105. (LCM of 7,6 and 7,15 respectively) But we need to add multiples of LCM of 7,6,15. since they would have been subtracted twice.

Following similar steps as above like we did for 7; no of multiples of 42 = 21, no of multiples of 105 = 9, no of multiples of 210 = 4

So answer is 128 - 21 - 9 + 4 = 102. answer is B.

nivivacious · Answer

We need 3-digit positive integers that are multiples of 7 but not multiples of 6 or 15
Multiples of 7 are 7, 14, 21, 28, 35, etc. 
-Smallest 3-digit number divisible by 7 = 105
-Largest 3-digit number divisible by 7 = 994
-994 = 105 + 7(n-1)  
-n = 128
Multiples of 6 are 6, 12, 18, 24, etc.
Multiples of 15 are 15, 30, 45, etc.

There are 26 such numbers
So we can rule out multiples of 6 and 15. The remaining multiples of 7 between 100 and 999 are 128-26=102
B

Kinshook · Answer

Asked: How many three-digit positive integers are multiples of 7 but not multiples of 6 or 15 ?

Number of 3-digit positive integers that are multiple of 7 = {105,....994} = (994-105)/7 + 1 = 889/7 + 1 = 127 + 1 = 128
Number of 3-digit positive integers that are multiple of (7 & 6 = 42) = {126,...,966} = (966-126)/42 + 1 = 840/42 + 1 = 21
Number of 3-digit positive integers that are multiple of (7 & 15 = 105) = {105,...., 945} = (945-105)/105 + 1 = 840/105 + 1 = 9
Number of 3-digit positive integers that are multiple of (7 & 30 = 210) = {210,...., 840} = (840-210)/210 + 1 = 630/210 + 1 = 4

Number of three-digit positive integers are multiples of 7 but not multiples of 6 or 15 = 128 - (21+9-4) = 128 - 26 = 102

IMO B

nj23598 · Answer

smallest 3-D number divisible by 7= 105,
total number of 3 digit nos. divisible by 7 are,
using AP formula we get, 994=105+(n-1)*7
n= 128.

Now, find out total numbers divisible by 6*7= 42 and 15*7= 105 separately.
we get,
for 42, 966=126+(n-1)*42
n=21.

for 105, 945=105+(n-1)*105
n=9.

Now take LCM of 6,15, we get= 30, now multiply 30*7=210,
Now calculate 3 digit numbers divisible by 210- no need for formula here, we can calculate manually- 210, 420, 630, 840.

Only divisible by 42 = 21
Only divisible by 105 =9
divisible by both 42 and 105 =4
Total= 21+(9-4)= 26,

Deduct 26 from 128 (3 digit numbers divisible by 7)
Hence, 128-26= 102.

Answer: 102 (B)

ARSarkar · Answer

7* 15 =105
7*142=994
 so total number of 7s multiples are (142-15)+1=128

Now we have to remove multiple of 6s, 15s and common of 6and 15.

7*18 is a multiple of 6.
6*3=18
6*23=138

total multiples of 6 are (23-3)+1=21 ......... (A)
 7*15 is a multiple of 6.
siilarly 7*30, 7*45, 7*60, 7*75, 7*90,7*105,7*120 and 7*135 are multiple of 6.
total multiple of 15 are 9. ... (B)

now 7*30, 7*60,7*90,7*120 are common. total common are 4. we ned to add these.

128-A-B+4=128-21-9+4=102.
Option B

cledgard · Answer

# multiples of 7 - # multiples of (7 U 6) - # multiples (7 U 15)   +   # multiples (7 U 6 U 15),  (multiples of 6 and 7 can concide) .
# of items =   + 1
# multiples of 7,   + 1 = 128
# multiples (7 U 6),   + 1 = 21
# multiples (7 U 15),   + 1 = 9
# multiples (7 U 6 U 15),   + 1 = 4
 # multiples of 7 - # multiples of (7 U 6) - # multiples (7 U 15) + # multiples (7 U 6 U 15) = 128 - 21 - 9 + 4 = 102 
Answer  102

B

sujoykrdatta · Answer

The first 3-digit multiple of 7 is 105 = 7 * 15
The last 3-digit multiple of 7 is 994 = 7 * 142 (dividing 999 by 7, we get the quotient and remainder; remove the remainder from 999; alternatively you know that 1001 is a multiple of 7)
So, the required multiples are 15, 16, 17 ... 142 => 142 - 15 + 1 = 128 numbers
We do not want the multiples of 6 or 15 
Number of multiples of 6 are: 7*18 ... 7*138 => (138 - 18)/6 + 1 = 21 numbers
Number of multiples of 15 are: 7*15 ... 7*135 => (135 - 15)/15 + 1 = 9 numbers
Number of multiples of 6 and 15, i.e. 30 are: 7*30, 7*60, 7*90 and 7*120 => 4 numbers
Thus, we need to remove 21 + 9 - 4 = 26 numbers from the 128 numbers

Required number = 128 - 26 = 102
Answer B

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