How many three digit natural numbers are divisible by 7?
Answers
Answered by
153
The first 3 digit number that is divisible by 7 is 105 and the last 3 digit number is divisible by 7 is 994.
To find out all the numbers falling between 105 and 994, that are divisible by 7 we will apply the formula AP= tn= a+ (n-1) d
Where a= 105, d=7, tn =994
Let number of terms= n, thus
994= 105+ (n-1) 7
994-105=7n-7
889=7n-7
n-1= 889/7
n-1= 127
n=127+1
n= 128
Therefore there are 128 three digit numbers that are divisible by 7.
To find out all the numbers falling between 105 and 994, that are divisible by 7 we will apply the formula AP= tn= a+ (n-1) d
Where a= 105, d=7, tn =994
Let number of terms= n, thus
994= 105+ (n-1) 7
994-105=7n-7
889=7n-7
n-1= 889/7
n-1= 127
n=127+1
n= 128
Therefore there are 128 three digit numbers that are divisible by 7.
Answered by
33
Answer:
n = 128
Step-by-step explanation:
The first 3 digit number that is divisible by 7 is 105 and the last 3 digit number is divisible by 7 is 994.
To find out all the numbers falling between 105 and 994, that are divisible by 7 we will apply the formula AP= tn= a+ (n-1) d
Where a= 105, d=7, tn =994
Let number of terms= n, thus
994= 105+ (n-1) 7
994-105=7n-7
889=7n-7
n-1= 889/7
n-1= 127
n=127+1
n= 128
Therefore there are 128 three digit numbers that are divisible by 7.
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