Apr 202020
Chris still has a pile of Easter eggs left.
- He can put the eggs exactly into piles of 7.
- If he puts them in piles of 3, 4, 5 or 6, he has one left over.
- What is the smallest number of eggs that Chris could have?
- * What is the next smallest number of eggs that Chris could have?
- ** What is the next, next smallest number of eggs that Chris could have?
15 Responses to “T5W1 Maths challenge”
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1- 7
2- 21
3- 28
Not quite George – 7 is one more than a multiple (piles) of 3 and 6, but not one more than a multiple (piles) of 4 or 5.
The smallest is 49 so the next smallest is 98 and the next next is 147
Not quite. 49 is one more that a multiple of 3, 4 and 6, but it is not one more than a multiple of 5.
301, because it goes into 7 and 3,4,5 and 6 go into 300.
301
301, because 7 goes into it and 3,4,5 and 6 go into 300.
The first one is 41
41 is not one more than a multile of 6 (or a multile of 7)
1st is 301
2nd is 721
3rd is 1141
The smallest number of eggs he could have is 301.
1st = 301
2nd = 721
3rd = 1141
301
721
1141
60 is the smallest number that 3, 4, 5 and 6 fit into exactly so answer needs to be 1 more than a number that 60 fits into exactly.
Fist =301
Second=721
Third=1141
60 is the lowest number 3,4,5and 6 fit into perfectly so the answer has to be one above a number that 60 fits perfectly into.
Well done to Mathhew and Jacob that got all three answers – 301, 721, 1141. And well done to everyone who got 301 as the first answer.
Here is a Scratch project that calculates the answers and puts them in a list. See if you can work out how it works… 31 mod 5 means the remainder when you divide 30 by 5 so “31 mod 5 = 1” so each of the IF lines is testing whether it is a multiple of 7 and if there is one left over for each of 6,5,4,3.
Easter egg maths problem scratch project