Take 500 and divide by 7. You're left with 71 with a remainder of 3. Now subtract 3 from 500 and you're left with 497. Divide 497 by 7 and you have 71 with no remainder. So 497 is the maximum nuber of eggs you can have in the basket.
2006-12-16 11:18:44
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answer #1
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answered by tropicalturbodave 5
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The number must be multiple of the LCM of 2, 3, 4, 5, & 6 +1
3*4*5=60
& must be a multiple of 7
possibilities are
61
121
181
241
301
... 301=7*43
so there were 301 eggs in the cart.
The naxt values that meete these criteria is
301+420=721 which is more than the cart will hold
2006-12-16 22:06:25
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answer #2
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answered by yupchagee 7
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there are 301 eggs in a cart... why?
you have to look first the common multiple of 2, 3, 4, 5 and 6...
common multiples of 2, 3, 4, 5, and 6 are 60, 120, 180, 240, 300, 360, 420 and 480..
add 1 to that multiple and try to divide it into 7.
300 is the only number that can answer to the problem...
301/2 = 150 r.1
301/3 = 100 r.1
301/4 = 72 r.1
301/5 = 60 r.1
301/6 = 50 r.1
301/7 = 43.. no remainder...
hope that this will help you...
2006-12-16 19:26:45
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answer #3
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answered by racz_jay25 2
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497
process of elimination starting at 500 and working your way backwards, divide by 2, 3, 4, 5, 6, 7 until you find a number that gives you the answer
2006-12-16 21:43:08
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answer #4
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answered by summer_00_butterfly 3
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there are 7 eggs in the cart
2006-12-16 19:15:03
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answer #5
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answered by stinkyhotdogs 3
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please report abuse for a guy with a username of "man"
he spams every single question he answers, yet he has 20,000 points. why has anyone ever reported him yet?! we should ALL report him and maybe y! answers will c the problem.
sry for crashing your question, but i wanted to get that thru.
2006-12-16 19:14:39
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answer #6
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answered by Anonymous
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do ur own dang homework
2006-12-16 19:15:37
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answer #7
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answered by JTTW9500 2
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lkkljl
2006-12-16 19:13:18
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answer #8
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answered by Anonymous
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please report abuse ''man''
2006-12-16 19:13:12
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answer #9
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answered by Anonymous
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