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If X and Y are the two digits of the number 347XY such that the number is completely divisible by 80, then what is the value of X + Y? (SSC CGL 2017)
- a)2
- b)4
- c)6
- d)8
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
If X and Y are the two digits of the number 347XY such that the number...
347XY as 347X0. Since 8 is a factor of 80.
347X0 is divisible by 8. It means last three digits 7X0 is divisible by 8.
Hence, X is 2 or 6
if X = 6, number is 34760. But this is not divisible by 80.
if X = 2, number is 34720, which is divisible by 80.
Therefore, number is 34720 with X = 2 and Y = 0.
∴ x + y = 2 + 0 = 2.
347X0 is divisible by 8. It means last three digits 7X0 is divisible by 8.
Hence, X is 2 or 6
if X = 6, number is 34760. But this is not divisible by 80.
if X = 2, number is 34720, which is divisible by 80.
Therefore, number is 34720 with X = 2 and Y = 0.
∴ x + y = 2 + 0 = 2.
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Community Answer
If X and Y are the two digits of the number 347XY such that the number...
To find the value of X and Y, we need to determine the conditions for the number 347XY to be divisible by 80.
Condition 1: Divisibility by 8
For a number to be divisible by 8, the last three digits of the number must form a multiple of 8. In this case, the last three digits are XY, so XY must be a multiple of 8.
Condition 2: Divisibility by 10
For a number to be divisible by 10, the last digit must be 0. However, in this case, the last digit is Y, so Y cannot be 0.
Condition 3: Divisibility by 80
For a number to be divisible by 80, it must satisfy both condition 1 and condition 2.
Now let's find the multiples of 8 and see which one satisfies condition 2.
Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, ...
We can see that the only multiple of 8 that does not end with 0 is 8 itself. So, Y must be equal to 8.
Now let's substitute Y = 8 into the number 347XY and check if it is divisible by 80.
347XY = 3478
Condition 1: Divisibility by 8
78 is a multiple of 8, so the number satisfies condition 1.
Condition 2: Divisibility by 10
The last digit is 8, which is not 0. The number does not satisfy condition 2.
Therefore, Y cannot be 8, and there is no solution that satisfies both conditions. Hence, the value of X and Y cannot be determined.
Condition 1: Divisibility by 8
For a number to be divisible by 8, the last three digits of the number must form a multiple of 8. In this case, the last three digits are XY, so XY must be a multiple of 8.
Condition 2: Divisibility by 10
For a number to be divisible by 10, the last digit must be 0. However, in this case, the last digit is Y, so Y cannot be 0.
Condition 3: Divisibility by 80
For a number to be divisible by 80, it must satisfy both condition 1 and condition 2.
Now let's find the multiples of 8 and see which one satisfies condition 2.
Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, ...
We can see that the only multiple of 8 that does not end with 0 is 8 itself. So, Y must be equal to 8.
Now let's substitute Y = 8 into the number 347XY and check if it is divisible by 80.
347XY = 3478
Condition 1: Divisibility by 8
78 is a multiple of 8, so the number satisfies condition 1.
Condition 2: Divisibility by 10
The last digit is 8, which is not 0. The number does not satisfy condition 2.
Therefore, Y cannot be 8, and there is no solution that satisfies both conditions. Hence, the value of X and Y cannot be determined.
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