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ok, for those who thought the last couple brain teasers were too easy, let’s kick it up a notch 
When asked how many nuts he had in his basket, a boy replied that when he counted them over 2 by 2, 3 by 3, 4 by 4, 5 by 5, or 6 by 6, there was 1 remaining; when he counted them by 7s there was no remainder. How many had he?
will post submitted solutions Monday, thanks
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301 nuts - I needed some help from Excel to solve this one but it worked nonetheless
I think it can be solved this way:
X= total of nuts the boy had
1) when you divided X by 2,3,4,5,6 there is a rest of 1
2) when you divided by seven there is no rest
Example: 11/ 2 = 5 (rest 1)= 2*5 +1= 11
So you can deduce that when the rest is 1 the result of the division is 1 number less, with this said lets go to the concrete bussiness:
1)X= 2A+1 (by 2)
2)X= 3B+1 ( by 3)
3)X= 4C+1 (by 4)
4)X= 5D+1 (by 5)
5)X= 6E+1 (by 6)
6)X= 7F (by 7 there is no rest)
so: ITS NOT DIVISIBLE BY 2,3,4,5,6 BUT BY 7 IS.
2A=3B=4C=5D=6E=7F-1 –> FROM HERE YOU DEDUCE THE FOLLOWING:
A has as factor 2, 5 and 3, B has 2 and 5, C (3,5), D(2,3) E(2,5), 7F-1(2,5,3)
X-1 HAS AS FACTORS 2,3,5–> 7F-1 HAS THE SAME FACTORS…
if its multiply by 5 minus 1, ends in 1 or in 6 but can not end in 6 because its not divisible by two, it ends in 1, it has to be a prime number too:
7*43=301—-> 301-1=300–>2(150)=3(100)=4(75)=5(60)=6(50)
the number is 301, i solved by induction, very hard this problem at least for me
in my previous comment i wanna clarify that the the one that i think has to be a prime number is the F not the X, just to make it more clear. It was hard but cool
721
ans. 301
m nt fully sure of the method ….but heres what i came out with…..
the number must be a multiple of 7 and a prime no.(as it dont get divided by 2,3,4,5,6)…..again since it gives remainder wid 5….so last digit must be 1 or 6…..7 gives these two last digit only when multiplied wid a num of last digit 3 or 8 respectively.here 8 can be eliminated(wont be prime)….thus it ends up to be either of these …7*13,7*23,7*43…..
here all the conditions are satisfied wid 7*43=301….thus i concluded the ans.
pls let me know if theres any fault in this method
ok, so we’ll need a number that can be factored by 7, and is one more than a number that can be factored by 6. the number must be odd, as it cannot be factored by 2. the digits my not add up to a number divisible by 3, as it could then be factored by 3. the number cannot end in 5.
ah, and since it is one more than a number factored by 5 and not even, the last digit MUST be 1.
getting warmer…
ugh, 7s are a pain to factor. it is the rythmless digit….
LCMs of 7 ending in 1 –{21…91…161…231} (learning something about the nature of 7s here. im close…)
there it is–next one, #301 is a factor of 7, and one less is a factor of 2,3,4,5 and 6.
301. final answer.
boo-ya.
301
2 * 3 * 4 * 5 * 6 + 1 = 721
He has 721 nuts in his basket
(2 x 3 x 4 x 5 x 6) + 1 = 721
(2 x 360) + 1 = 721
(3 x 240) + 1 = 721
(4 x 180) + 1 = 721
(5 x 144) + 1 = 721
(6 x 120) + 1 = 721
7 x 103 = 721
X is the number of nuts.
X - 1 is the number with which we will work, from the standpoint of 2, 3, 4, 5, and 6.
Since 2, 3, 4, 5, and 6 have no remainders when taken against X - 1, they are all factors of X - 1. Since 2 and 3 are factors for 6, 6 covers them. A factor set for 4 is 2*2. One of these 2’s takes care of the 2 in 6. 5 is the final factor. So the full factor set for X - 1 is 4*5*3, which = 60. So the number of nuts to meet this criterion must be an integer multiple of 60. Since 7 is a factor of X, the final result must be (n*60 + 1)/7 with no remainder. The first such value for which this true is 301.
A couple of 18 pack cartons of eggs (36 eggs):)
I’m nuts!!! I meant nuts, not eggs…
Three hunderd and one (301) nuts.
301
301
301 nuts
you could look at as 7 x a prime number (43)
or 2,3,4,5,6 intersect every 60 nuts, count by 60’s until the total +1 divides by 7 exactly
2×3x4×5x6 = 720
720 + 1 = 721
721 / 7 = 103
therefore he had 721 nuts in his basket
Number 49
Persistence & brainache produced 301!
Couldn’t be an even number. Had to be divisable by 7. Had to end in a 1 or 6 (because divisable by 5 with 1 left over), so just plodded my way through the 7 times table! Would be interested in knowing if there was a more logical way of working it out…
301
301, by brute force
The boy has 301 nuts in his basket
721 works, but so does 301!
Doug provides a nice explanation, as does Suineg, Kevin from Bathurst, John Lewis Highsmith, & bubbly and Tori too.
I think there is a relationship beteween the two answers:
1) 301 = 7*43, 721=7*103
2) 43*7+60*n*7 its a series that gives you all numbers that follow the rule, explanation:
43 its the base number, the minimal prime number that allows the rule to begin, 60= 2*2*3*5(minimal common factors)–> 2*3=6 Or 2*2=4, the shared 1 two, you move in the prime numbers following the root(43)–>examples:
43*7 +60*7=7*103–>321
43*7 +120*7=7*163–>1141
and so on….
In my explanation for this puzzle “integer multiples” should be replaced by “whole number multiples”. since I suspect that negative numbers would not lead to a satisfactory solution.
Another way of seeing the series of prime numbers is this:
43= 13 +30—-> 30=(2*3*5) the prime numbers involved as factors
103= (30+13)+(30+30)
163=(30+30+13)+(30+30+30)
and so on… you have to add 30 to both terms so that the result can be divisible to the factors, and the 13= 5*2+3 close the deal because it had the 2 that is need to have the 6 and the 4.
Cool, I hoped my reasoning its good, if there is some flaw in it, please let me know.
there absolutely appears to be a relationship. i had not considered the fact that the number multiplied by 7 had to be prime until i read some of these comments…because if it is multiplied by a number that is NOT prime, it would have to be a multiple of 2-6…..i think….
so i had the theory that a number that was a perfect square of seven that ended in a 1 would also work, and sure enough, 7 to the 4th power, or 2401, is also a valid answer. and 7 to the 8th power (5,764,901) also works. i suspect that 7^12, 7^16, 7^20 and so on will all satisfy this problem.
HOWEVER, the question said that a boy was counting the nuts, therefore 5,764,901 would NOT be a valid answer because no boy could count that many nuts, let alone have that many in a basket. so i say 301, and 721 are the only valid answers.
also, by doing bad math in my head while trying to figure out what 7^4 was (i added 700 to 1400 to get 14700)–hey, i was tired– i found that 15001 also works, and like 301, is not a perfect square of seven. if someone could break down formula that will always be true for this problem in a concise way, id like to see it.
7s arent so bad after-all. definetly unique.
Doug I think the formula I post it work, there is a flaw in it??:
its like this:
N is a positive integer.
Formula: 43*7+60*N*7
Explanation:43 is the root of the series–> 43= 13->(2*5)+3 (+)
30->(2*3*5)-> all prime numbers involved but you need another 2 to get the factor (4) that you get from the 13.
then you add 30(2*5*3) to both members and go on to infinity.
examples:
1)(13+30)*7-> 43*7=301-> root of the series N=0
2)((13+30)+(30+30))*7->(43+60)*7->103*7=721
3) ((13+30+30)+(30+30+30))*7->(73+90)*7->163*7=1141
1141-1=1140/2=570;1140/3=380;1140/4=285;1140/5=228;1140/6=190
and so on…
511?
Five hundred ten is divisible by 2,3,4,5, and six while always having one left over, except when you divide it by seven. So I suppose it is a plausible answer, but I’m not sure about the lowest possible.
LCM (2,3,4,5,6) = 60
60x + 1 = 7y where x and y are whole numbers
x = 7n + 5, where n = 0, 1, 2, 3…
The answer is 301 or this number plus any multiple of 420.
Such as 301, 721, 1141, 1561 and so on.
Basically because the LCM of 2,3,4,5,6 is 60, and 60×7=420.
60/7 has a remainder of 4.
What multiple of 4 is 1 less than a multiple of seven?
5
5×60=300
300+1=301
Any multiple of 301
A simple induction logic.
We’ll take 2,3,4,5. We’ll ignore 6 as 6 = 2 * 3
Let the no. of nuts = X
X%2 = X% 3 = X%4 = X%5 = 1
X should end with 1 or 6 to satisfy X%5
S should be odd to satisfy X%2
considering both, X should end with 1.
So, X should be a multiple of 7 which ends with 1 -i.e., 21,91,161,231,301,etc
301 satisfies X%2 = X% 3 = X%4 = X%5 = 1
So, the number of nuts should be a multiple of 301
Actually, a solution exists for every x where 7*(43+x60).
x=0, 301
x=1, 721
x=2, 1,141
There are infinitely many solutions!!!
How big a nut basket do ya got?