
Given that the number $\overline{67y19}$ is divisible by 9, where y is a digit, what are the possible values of y?
Answer
508.5k+ views
Hint: We solve this problem by first discussing the divisibility rule for 9 and then use it to see under which conditions the given number $\overline{67y19}$ is divisible by 9. Then we substitute the values from 0 to 9 and check if it satisfies the obtained condition. The values that satisfy the condition are our required values.
Complete step by step answer:
Before solving the problem let us go through the divisibility rule for number 9. Any number is said to be divisible by 9 if the sum of all the digits of the number is divisible by 9.
For example, let us take two numbers 34567 and 24795.
Now let us see the first number 34567.
Sum of its digits is 3+4+5+6+7=25. As we see 25 is not divisible by 9. So, the number 34567 is not divisible by 9.
Now let us take a look at the second number, 24795.
Sum of its digits is 2+4+7+9+5=27. As 27 is divisible by 9, the given number 24795 is divisible by 9.
We are given that the number $\overline{67y19}$ is divisible by 9.
So, by using the divisibility rule of 9, the sum of the digits must be divisible by 9.
Sum of the digits is equal to $6+7+y+1+9=y+23$.
So, y+23 needs to be divisible by 9. As y is a digit it takes values from 0 to 9. Let us see which of those values satisfies our condition.
When y=0, $y+23=0+23=23$, not divisible by 9.
When y=1, $y+23=1+23=24$, not divisible by 9.
When y=2, $y+23=2+23=25$, not divisible by 9.
When y=3, $y+23=3+23=26$, not divisible by 9.
When y=4, $y+23=4+23=27$, is divisible by 9.
When y=5, $y+23=5+23=28$, not divisible by 9.
When y=6, $y+23=6+23=29$, not divisible by 9.
When y=0, $y+23=7+23=30$, not divisible by 9.
When y=0, $y+23=8+23=31$, not divisible by 9.
When y=0, $y+23=9+23=32$, not divisible by 9.
So, the sum of the digits is divisible by 9, when y=4. So, the given number is divisible by 9 when y=4.
So, the correct answer is 4.
Note: We can also solve this problem by substituting the values 0, 1, 2, 3,….., 9 in y and checking if the number $\overline{67y19}$ is divisible by 9 by dividing the number using long division as
But it is a hectic way of solving the problem. So, it is better to remember the divisibility rule for 9 and use it for solving this type of question.
Complete step by step answer:
Before solving the problem let us go through the divisibility rule for number 9. Any number is said to be divisible by 9 if the sum of all the digits of the number is divisible by 9.
For example, let us take two numbers 34567 and 24795.
Now let us see the first number 34567.
Sum of its digits is 3+4+5+6+7=25. As we see 25 is not divisible by 9. So, the number 34567 is not divisible by 9.
Now let us take a look at the second number, 24795.
Sum of its digits is 2+4+7+9+5=27. As 27 is divisible by 9, the given number 24795 is divisible by 9.
We are given that the number $\overline{67y19}$ is divisible by 9.
So, by using the divisibility rule of 9, the sum of the digits must be divisible by 9.
Sum of the digits is equal to $6+7+y+1+9=y+23$.
So, y+23 needs to be divisible by 9. As y is a digit it takes values from 0 to 9. Let us see which of those values satisfies our condition.
When y=0, $y+23=0+23=23$, not divisible by 9.
When y=1, $y+23=1+23=24$, not divisible by 9.
When y=2, $y+23=2+23=25$, not divisible by 9.
When y=3, $y+23=3+23=26$, not divisible by 9.
When y=4, $y+23=4+23=27$, is divisible by 9.
When y=5, $y+23=5+23=28$, not divisible by 9.
When y=6, $y+23=6+23=29$, not divisible by 9.
When y=0, $y+23=7+23=30$, not divisible by 9.
When y=0, $y+23=8+23=31$, not divisible by 9.
When y=0, $y+23=9+23=32$, not divisible by 9.
So, the sum of the digits is divisible by 9, when y=4. So, the given number is divisible by 9 when y=4.
So, the correct answer is 4.
Note: We can also solve this problem by substituting the values 0, 1, 2, 3,….., 9 in y and checking if the number $\overline{67y19}$ is divisible by 9 by dividing the number using long division as

But it is a hectic way of solving the problem. So, it is better to remember the divisibility rule for 9 and use it for solving this type of question.
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