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What digit must be assigned to k so that the 7-digit number 451k603 is exactly divisible by 9?
5
8
7
2
8
A number is divisible by 9 if the sum of its digits is a multiple of 9. Quickly sum the known digits and find what must be added to reach the next multiple of 9.
The 7-digit number is 451k603 and it must be divisible by 9.
Sum of digits=4+5+1+k+6+0+3
A number is divisible by 9 if the sum of its digits is a multiple of 9. Quickly sum the known digits and find what must be added to reach the next multiple of 9.
Confusing the divisibility rule of 3 with 9, or making arithmetic errors in the summation of the digits.
State the divisibility rule for 9
A number is exactly divisible by 9 if and only if the sum of all its digits is divisible by 9.
4+5+1+k+6+0+3=19+k
Find the next multiple of 9
We need the smallest digit k (where 0≤k≤9) such that 19+k is a multiple of 9. The multiples of 9 are 9, 18, 27, 36, etc.
19+k=27
Solve for k
Subtracting 19 from 27 gives the value of k.
k=27−19=8
B is correct because the sum of digits 19+k requires k=8 to become 27, which is divisible by 9.
This same 'Sum of Digits' rule applies to divisibility by 3, but the divisor is 3 instead of 9; always check the divisor carefully in exams.
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