+0

0
166
8

If counted in pairs, 1 remain

If counted in threes, 2 remain

If counted in fours, 3 remain

If counted in fives, 4 remain

If counted in sixes,5 remain

If counted in sevens, none remain

His basket cannot accomodate more than 150 eggs

Jun 23, 2020

#1
+1

so say t is total eggs

t<=150

t is odd

2 more than multiple of 3, or 1 less than multiple of 3

3 more than multiple of 4, or 1 less than of 4

4 more than of 5, or 1 less

5 more than of 6, or 1 less

Multiple of 7

multiples of 7 to 150

7,14,21,28,35,42,49,56,63,70,77,84,91,98,105,112,119,126,133,140,147

Two more than multiples of 3

21-2 = 19, no

35-2=33, huh.

49-2 = 47, no

63-2=61, no

77-2 = 75, huh.

91-2 = 89, no

105-2=103, no

119-2 = 117, huh.

Rest are no's.

35, 77, 119

35-3= 32, huh.

77-3 = 75, huh.

119-3 = 116, huh.

35-4 = 31, no.

77 - 4 = 73, no

119-4 = 115, yes.

119.

Just to check, 119-5 = 114,/6 = 19, yes, and it's a multiple of 7,

Apologies for my prior mistake and thanks for bringing this to my attention, please check to ensure correctness.

Jun 23, 2020
edited by hugomimihu  Jun 23, 2020
#2
0

This isn’t the correct answer Hugo! You didn’t compare your result to the criteria that defines the problem.

If you did compare you would see this:

If 75 eggs are counted in threes, then 0 remain; it should have 2

If 75 eggs are counted in fives, then 0 remain; it should have 4

If 75 eggs are counted in sixes, then 3 remain; it should have 5

And the most obvious one:

If 75 eggs are counted in sevens, then 5 remain; it should have 0

Here’s a hint: Reread the question, and notice that the remainders are all one less than the divisor, except for seven.

It’s not likely you will know what to do with this hint. So here are few more hints.

What do you get by taking the LCM of the divisors that have a remainder other than zero?

What happens when 1 is subtracted from this LCM?

Is the result divisible by 7?

If not, then what about  2*(LCM - 1)?

.
Jun 23, 2020
edited by Guest  Jun 23, 2020
#6
0

hugomimihu  Jun 23, 2020
#3
+1

N mod  2 = 1

N mod 3 = 2

N mod 4 = 3

N mod 5 =4

N mod 6 =5

N mod 7 =0

Using Chinese Remainder Theorem plus Modular Multiplicative Inverse which are incorporated in this short computer code, we get the following:

i=0;j=0;m=0;t=0;a=(2, 3, 4, 5, 6, 7);r= (1, 2, 3, 4, 5, 0);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return

OUTPUT =420n  +  119, where n =0,  1, 2, 3........etc.

Therefore, the smallest number of eggs he had was = 119 eggs.

Jun 23, 2020