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Secondary 1 | Maths
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Date Posted: 4 years ago
Views: 223
8x16y is greater than 82000, so x can only be 2, 3, 4, 5, 6, 7, 8 or 9.
Since the number is divisible by 5, the final digit must be 5 or 0.
Since the number is divisible by 3 as well, the sum of the 5 digits must be divisible by 3.
If x is 2, then the sum of all the digits except y is 8 + 2 + 1 + 6 = 17. Adding a 0 or 5 will yield a number which is not divisible by 3. So, x cannot be 2.
If x is 3, then the sum of all the digits except y is 8 + 3 + 1 + 6 = 18. This is already divisible by 3, so y can be 0.
So, the smallest value of 8x16y is 83 160.
Since the number is divisible by 5, the final digit must be 5 or 0.
Since the number is divisible by 3 as well, the sum of the 5 digits must be divisible by 3.
If x is 2, then the sum of all the digits except y is 8 + 2 + 1 + 6 = 17. Adding a 0 or 5 will yield a number which is not divisible by 3. So, x cannot be 2.
If x is 3, then the sum of all the digits except y is 8 + 3 + 1 + 6 = 18. This is already divisible by 3, so y can be 0.
So, the smallest value of 8x16y is 83 160.
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Use divisibility rules of 5 and 3, list down possible combinations.
Date Posted:
4 years ago
Thank you
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