Q&A - P4 Math
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quekcc:
18 coins in piles of 5 will have 3 coins left over, not 2.
1. The number n is a multiple of 6.Bunny27:
Hi,
Can someone kindly help me on the question ?
1. Chenli has a collection of coins. When she put the coins in piles of fours or fives, she has 2 coins left over each time. When she puts them in piles of 6s, she has no coins left over. What is the smallest number of coins that Chenli can have?
Thanks!
2. The number (n + 2) is a multiple of 4 and 5, i.e. it must end with 0.
3. implies that n must end with 8.
Smallest multiple of 6 ended with 8 will be 18.
Ans: 18 coins.
Condition 2 should be (n - 2) is multiple of both 4 & 5, so n must end with \"2\".
Answer: 42 coins -
BigDevil,
Condition 2 should be (n - 2) is multiple of both 4 & 5, so n must end with "2". Why must end with 2?
Why not n+2 since 2 coins left over each time? -
How to do this question?
Mr Tong bought some candy bars. If shard by 3 children, there was no remainder left, but when shared the candy bars by 5 children, there were 3 candy left over. What is the min no. Of candy bars Mr To g bought? -
cathycho:
the \"standard\" approach is the list the multiple of 3 and multiple of 5+3 and look for the smallest number that matches.How to do this question?
Mr Tong bought some candy bars. If shard by 3 children, there was no remainder left, but when shared the candy bars by 5 children, there were 3 candy left over. What is the min no. Of candy bars Mr To g bought?
multiples of 3 --> 3, 6, 9, 12, 15, 18, 21....
multiples of 5 + 3 --> 8, 13, 18, 23....
alternate approach:
If we remove 3 candy bars, the remaining number will be divisible by 3 & 5.
The smallest number that is divisible by 3 & 5 is 15.
The number of candy bars = 15 + 3 =18
cheers. -
MathIzzzFun,
Thanks for ur help. :lovesite: -
cathycho:
Let's recap what is said in the question...BigDevil,
Condition 2 should be (n - 2) is multiple of both 4 & 5, so n must end with \"2\". Why must end with 2?
Why not n+2 since 2 coins left over each time?
When she put the coins in piles of fours or fives, she has 2 coins left over each time.
When the number n is divided by 4 or 5, there is a remainder of 2.
If we take 2 away from n, then the end number is divisible by 4 and 5.
Thus, (n - 2) is multiple of both 4 and 5, not (n + 2).
For (n -2) to be divisible by 5, it must end with 5 or 0.
But any number ending with 5 cannot be a multiple of 4,
so (n -2) must end with 0. Meaning n must end with 2.
HTH. -
Poppet:
Jeremy has some sweets.If he puts 4 sweets in each bag,there will be 15 sweets left over.He puts 6 sweets in each bag,there will be 3 sweets left over.How many sweets does Jeremy have?
Can someone help? :xedfingers:gonzo:
27 is certainly a valid answer to the question, but what troubled me is that if Jeremy still has 15 sweets left over, why didn't he keep filling more bags of 4 till he has only 3 left?Multiples of 4: 4,8,12,16,20
Add 15: 19, 23, 27, 31, 35
Multiples of 6: 6,12,18,24,30
Add 3: 9,15, 21,27,33
Common number is 27.
So I'm tempted to think that he has only a fixed number of bags.
Filling the bags with 4 sweets, he has 15 left.
Filling the bags with 6 sweets, he has 3 left.
So to put in 2 extra sweets into each bag, he used up 12 additional sweets.
Therefore, he has 12/2 = 6 bags.
Total number of sweets: (6 x 6) + 3 = 39 -
BigDevil:
Finally, I understand! Thks for ur explanation..
Let's recap what is said in the question...cathycho:
BigDevil,
Condition 2 should be (n - 2) is multiple of both 4 & 5, so n must end with \"2\". Why must end with 2?
Why not n+2 since 2 coins left over each time?
When she put the coins in piles of fours or fives, she has 2 coins left over each time.
When the number n is divided by 4 or 5, there is a remainder of 2.
If we take 2 away from n, then the end number is divisible by 4 and 5.
Thus, (n - 2) is multiple of both 4 and 5, not (n + 2).
For (n -2) to be divisible by 5, it must end with 5 or 0.
But any number ending with 5 cannot be a multiple of 4,
so (n -2) must end with 0. Meaning n must end with 2.
HTH. -
BigDevil:
I agree with you. A similar question was discussed quite some time ago- http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=66&t=149&hilit=lim+has+some+sweets&start=1030.Poppet:
Jeremy has some sweets.If he puts 4 sweets in each bag,there will be 15 sweets left over.He puts 6 sweets in each bag,there will be 3 sweets left over.How many sweets does Jeremy have?
Can someone help? :xedfingers:gonzo:
27 is certainly a valid answer to the question, but what troubled me is that if Jeremy still has 15 sweets left over, why didn't he keep filling more bags of 4 till he has only 3 left?Multiples of 4: 4,8,12,16,20
Add 15: 19, 23, 27, 31, 35
Multiples of 6: 6,12,18,24,30
Add 3: 9,15, 21,27,33
Common number is 27.
So I'm tempted to think that he has only a fixed number of bags.
Filling the bags with 4 sweets, he has 15 left.
Filling the bags with 6 sweets, he has 3 left.
So to put in 2 extra sweets into each bag, he used up 12 additional sweets.
Therefore, he has 12/2 = 6 bags.
Total number of sweets: (6 x 6) + 3 = 39
cheers. -
Thanks, MathIzzzFun.
Oh my...that was discussed as a P3 question??
Seems a bit too tough for P3...don't think my P4 DD would know how to solve this!
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