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8a) Yes, but only under the circumstances where m & n are multiples of 2 (even numbers), hence it is possible to divide both m & n by 2 completely and form a whole number of 2cm x 2cm squares with no remainder.

8b) To find the number of ways to split this rectangle into squares, we need to find the number of common factors between 280 and 145:

280=1x2x2x2x5x7

145=1x5x29

As such these 2 numbers only have 2 commin factors, 1 and 5, and it is only possible to split this rectangle into squares in 2 ways.

8b) To find the number of ways to split this rectangle into squares, we need to find the number of common factors between 280 and 145:

280=1x2x2x2x5x7

145=1x5x29

As such these 2 numbers only have 2 commin factors, 1 and 5, and it is only possible to split this rectangle into squares in 2 ways.

Thank you soo much

1176 ÷ 4 = 294

294 can be further divided by 2

294 ÷ 2 = 147

147 can be further divided by 3, as the sum of its digits are a multiple of 3:

147÷3=49

49 also equal to 49=7x7

as such 1176=4x2x3x7x7=2³x3¹x7²

hence, a=3, b=1 and c=2

9bi)

1176=2x2x2x3x7x7

if the HCF of 1176 and m is 2x2x3x7

Since m is a multiple of 2x2x3x7, and that is already a factor of 1176, then Smallest Value of M=2x2x3x7

9bii) For another possible value of M, it will have a to be a multiple if 2x2x3x7 but have a factor that is not one of the prime factors of 1176 (2, 3 or 7). For example 5, or 11, or 13, as such, m can also be

a) m=5x2x2x3x7

b) m=11x2x2x3x7

c) m=13x2x2x3x7