Given that the number 67y19 is a multiple of 9,where 'y' is a digit,what are the possible values of 'y'
Answers
Answer:
4
Step-by-step explanation:According to divisibility rule of 9 is that sum of the digits must divisible by 9.
The number is 67y19, where 'y' is a digit.
Now,find the sum of the above number: 6+7+y+1+9=24+y.
So,the least value of y will be 3 ,so that 24+3 become 27 which is divisible by 9.
Therefore,the possible value of 'y' is 3.
Answer:
4
Step-by-step explanation:
Since 67y19 is a multiple of 9, the sum of digits will be divisible by 9.
Sum of the digits = 6 + 7 + y + 1 + 9
(6+7+y+1+9)/9 = k/1
23 + y = 9k ----> 1
Let k = 1,
23 + y = 9k
23 + y = 9×1
23 + y = 9
y = 9 - 23
= -14 not possible
Let k = 2,
23 + y = 9k
23 + y = 9×2
23 + y = 18
y = 18 - 23
= -5 not possible
Let k = 3,
23 + y = 9k
23 + y = 9×3
23 + y = 27
y = 27 - 23
= 4
Put y = 4 in 67y19 = 67419
Sum of the digits = 6 + 7 + 4 + 1 + 9 = 27 is a multiple of 9.
One possible value of y is 4
Let k = 4,
23 + y = 9k
23 + y = 9×4
23 + y = 36
y = 36 - 23
= 13 which is not a digit
which is not a digit So, not possible
Hence the required value of y is 4.
Hope this helps you !!!