Q34 How many 3 digit positive integers exist that when divided by 7 leave a remainder of 0?
Answers
Answer:
128 numbers are didvisble by 7 that leave remainder 0
Step-by-step explanation:
Let a be the first 3-digit no. divisible by 7
To get a, divide 100 by 7. We get,
100=(7×14)+2
We know that 7×14 is a multiple of 7 and 100 is not. Since 100 is not a multiple of 7
a=100 +5 =105
[ 7×14 is 98 and next multiple of 7 is 105. 100 is only 2 times more than 98. To get the next multiple of 7 we have to add 5 to 100]
Let l be the last 3 digit no. divisible by 7
To get l we divide l by 999. We get 5 as the remainder. We subtract 5 from 999 to get l.
l= 999-5=994
We know that l=a+(n-1)d
Where, n is total no. of 3-digit numbers divisible by 7 and d is the common difference between successive numbers, which is 7
Therefore,
l=a+(n-1)d
994=105+(n-1)7
889=(n-1)7
889/7=n-1
n-1=127
n=128
Thus there are 128 3-digit numbers divisible by 7.
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