What number divided by 2, 3, 4, and 5 has a remainder of 1, but when divided by 7 has no remainder?

Question

This is a math logic problem posed in my daughter's 5th grade math class. Sometimes these have fairly straightforward math solutions and other times they are more like a riddle. We're stuck!!

Answer 1

This question has been posed before but I'll give a method
of solving it. It's not really a riddle.
Let's work it one part at a time.
If the number is of the form 4t +1, it is automatically
of the form 2t +1.
Let x = 4a + 1
and x = 3b + 1.
Then 4a = 3b, which means a is a multiple of 3,
because 3 divides 4a and 3 doesn't divide 4.
Since 3 is prime, it must divide a.
a = 3c.
So x = 12c +1.
Now x = 5d + 1
and 12c = 5d,
so c is a multiple of 5.(Same reason as before)
Now x = 60d + 1
and x= 7f
So 7f = 60d + 1
So 7 must divide 60d +1
The smallest d for which this happens is d = 5
so the smallest number satisfying all the
conditions of the problem is 301. It turns out that
all the numbers that satisfy 7 dividing
60d +1 are of the form 7t + 5.
So x = 60(7t + 5) + 1 = 301 + 420t.
So the next one is 301 + 420 = 721.
This procedure is an example of the Chinese
remainder theorem. A version of it was known
in China around 1100 B.C.!

Answer 2

The LCM of 2, 3, 4 and 5, plus one, will have a remainder of 1 when divided by each of these numbers. That is 61, as the previous answerer said.

This will also be true for any multiple of 60 plus 1:
61, 121, 181, 241, etc.

But only multiples of 7 will do. As the previous answerer said, the first number that will work is 301. If you add 7x60 = 420 to this answer you can generate more solutions:
301+420 = 721
...+420 = 1141
...+420 = 1561
...+420 = 1981
...+420 = 2401
etc.

Answer 3

this is straighforward logic. it must be a multiple of 7. since the remainder must be one when divided by 2 and 5, the last number must be a 1. (if you use 56 for example and divide by 2 the remainder is not 1) the second to last digit must be an even number or it will not fit with the 4. the added digits must be one greater than a multiple of 3, (all digits divisible by 3 have digits that add up to a multiple of 3 eg 12, 1+2=3, 54, 5+4=9 etc) now the least number that is divisible by 2, 3 ,4 and 5 is 60. so this number will be 1 greater than a number divisible by 60. and it will also be divisible by 7. by now we have so many guidelines that it is easy. lets just multiply numbers by 60 and add 1 until we hit it. 60 x 2 wont work because it isnt divisible by 7.
60 x 3 doesnt work because it isnt divisible by 7. 60 x 4 doesnt work. 60 x 5 = 300 + 1 = 301. this is the first number that works

so the answer is 301

if i helped vote mine best :D

Answer 4

It's divisible by 7, but not divisible by 2,3,4, or 5, and specifically leaves a remainder of 1 when divided by those numbers.

Because it is not divisible by 2, we know that it has to be an odd number, and because it leaves a remainder of 1 when divided by 5, it has to end in 1 or 6 (because all multiples of 5 end in 0 or 5). These two facts combined let us know that it has to end in 1 (and not 6, which is divisible by 2).

So, we can look at multiples of 7 that don't also have factors of 2,3,4 or 5. that have the last digit of 1.
for 7 x anything to end in 1, the other last digit must be a 3.
7 x 13 = 91 but 91 / 4 has a remainder of 3.
7 x 23 = 161 but 161 / 3 has a remainder of 2
7 x 33 is obvious divisible by 3
7 x 43 = 301. which does leave a remainder of 1 when you divide it by 2,3,4, or 5. 301 is the lowest number for which this puzzle is true.

Opening up a spreadsheet to help me out, I find that 721 also works, as do:
1,141
1,561,
1,981
2,401
2,821
3,241.. etc..
basically any number in the format 301 + 420*x where x=0,1,2,3,4....

if you want to wow the teacher, use 108,661
it is 7 x 15,523.

Answer 5

the nohas to be a multiple of 120 plus 1
720+1 fits in
as 720 being a multiple of the LCM of 2,3,4,5 will leave a remainder of 1 when increased by 1
and 721/7=103

Answer 6

2: it's odd, since it is equal to 2x+1
and 3x+1
and 4x+1
and 5x + 1
(so, the number -1 is a multiple of all of these)
3*4*5+1 = 61 works and I believe it is the smallest number that satisfies the initial conditions, but that is not divisible by 7.
61
so we need to find m so that 60*m+1 is dividible by 7.
301 works.
I have to admit, though, that I did this by trial and error

Answer 7

721