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Secondary 1 | Maths
One Answer Below
Anyone can contribute an answer, even non-tutors.
Plz advise
Date Posted: 1 year ago
Views: 205
131A2B's value
= 131 000 + A × 100 + 2 × 10 + B × 1
= 131 020 + 100A + B
If we want the smallest possible number, then let A = 0
This means that 131020 + B must be a multiple of 15.
Now using long division, 131020 ÷ 15
= 8734 R10
So to get a multiple of 15, we have to let B = 5 such that there is no remainder.
(The 10 combines with the 5 to become 15)
Smallest possible = 131025
= 131 000 + A × 100 + 2 × 10 + B × 1
= 131 020 + 100A + B
If we want the smallest possible number, then let A = 0
This means that 131020 + B must be a multiple of 15.
Now using long division, 131020 ÷ 15
= 8734 R10
So to get a multiple of 15, we have to let B = 5 such that there is no remainder.
(The 10 combines with the 5 to become 15)
Smallest possible = 131025
See 1 Answer
131025
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