Two Digit > 9 Remainder 6 8, 16, 24 14, 22, 30 9, 18, 27, 12, 21, 30, Answer is 30 Three Digit > 99 96, 104, 112, 120 102, 110, 118, 126 99, 108, 117 102, 111, 120 Answer is 102

Remember: 1000 is a product of 8 while 999 is a product of 9. So From here, list down the possible numbers. 998 996 990 987 982 976 974 968 966 979 978 970 970 …… Hence, the number is 970.

The left column are the numbers that give a remainder of 6 when divided by 8, while the right column are the numbers that give a remainder of 3 when divided by 9.

is it possible to have a formula, in case the listing of these combination is long?

Zhong Shu Hao the right column doesn't give remainder of 3 when divided by 9

oh... oops i will correct it

Remember: 1000 is a product of 8 while 999 is a product of 9. So From here, list down the possible numbers. 998 993 990 984 982 975 974 966 966 …… Hence, the number is 966.

er... by observation I guess... As you can see, the left column starts from the number 998 while the left column starts from 993, which is 5 less. But as the right column decreases at a greater rate, we will try to make the number in the left column SMALLER than the number in the right column, and the numbrr will be 990, which is 3 less than the number in the right column. Since the right column -9 each time and the left column -8 each time, the right column -1 extra than the left column each time. To end up with the same numbers in both columns, the right column must minus 3x9 while the left column must minus 3x8. 993 - 3x9 ＝993-27 ＝ 966 990 - 3x8 ＝990-24 ＝ 966

My bad, greatest number for part b) not smallest. As for a formula, not really. Besides this is a P4 question (of course can be given to P5 and P6) however this is introduced at P4 as part of factors and multiples and thus a formula would not be provided. Typically, the questions will be designed such that the listing does not need to be too long.

Primary sch Q tests quite a bit on observation skills... whether in this kind of questioms or models, or patterns (needless to say :D )

Thanks for your input :-)

Btw guys, this question came out from a p3 assessment book.

A similar question was posted not long ago.... https://www.facebook.com/photo.php?fbid=1447643575547894&set=gm.461776310645697&type=1

Hi there. This is pretty much the same as this question https://www.facebook.com/photo.php?fbid=10153333265138708&set=p.10153333265138708&type=1&theater

This original question is here https://www.facebook.com/photo.php?fbid=10205091566463649&set=gm.461028260720502&type=1&theater

Thx

What we can realise is that if we add 40, 56 and 60 together, the value we get is 2 times of the sum of what the 3 boys have altogether.

Jimmy + Muthu (56) plus Peifen + Muthu (60) = Peifen + Jimmy + Muthu + Muthu = 116. Since Peifen + jimmy = 40, Muthu + Muthu = 116-40, Muthu = 76/2 = 38.

Thank you so much for the solutions. Now it makes sense.

Suppose all are man. The amount of donated will be 13x$8=$104. But the actual amount of money donated is $139, which is $35 more. This $35 is due to the $7 extra that each woman donates. Hence, there will be $35/$7 = 5 women. Thus, there are 13-5=8 men.