There is no number that divides by 1 and leaves any remainder but 0.
But, if we exclude 1 from the list, 301 might be one answer. 721 is another. There are many more.
The least common multiple of 2, 3, 4, 5, 6 is 60. Therefore, (any multiple of 60) + 1 will work: 61, 121, 181, 241, etc.
Answer: 1081
we are solving the system of equations, well a system of linear congruences actually.
x=1 mod2
x=1 mod3
x=1 mod4
x=1 mod5
So we can use the Chinese Remainder Theorem. I won't include all the work but you get
Solutions: 1, 61 (mod 120)
Now you want a 4 digit number, so we use the fact that the LCM of 2,3,4 and 5 is 60
and add 60 however many times we need to get a 4 digit number
let's add 60x17=1020
the add 61 and we have 1081
Now let's check
if we divide any odd number by 2 remainder is 1
if we divide our number by 3 we have 360 so remander is 1
if we divide by 4 we have 270 with remainder 1
if we divide by 5 we have remainder 1 since those number divisible by 5 end in 0 or 5 and 1081 is 1 more than that.
so our answer works!
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or
Just use the Least Common Denominator for 2, 3, 4, and 5 and select the lowest multiple that has 4 digits. The LCD is 3 x 4 x 5 =60, which means that any number 1 more than a multiple of 60 will give you the above results. The first multiples of 60 with 4 digits are 1020 and 1080.
Both 1021 and 1081 will give you the remainder 1 for each division. So will any number that is 1 more than a 4 digit multiple of 60 (such as 6001).
1021 / 2 = 510 r. 1
1021 / 3 = 340 r. 1
1021 / 4 = 255 r. 1
1021 / 5 = 204 r. 1
61,121,181,241,301,361,421,481,541,601,661,721,781,841,901,961,1021,1081,1141,1201, ect
301: 7*43. 300 is divisible by 2, 3, 4, 5, and 6.
119
27
12
8
Every 2 digit number, when divided by the number one less than it, will result in a remainder of 1.
A 2 digit number divided by a four digit number, such as 2345, will leave the whole 2-digit number as a remainder. It cannot leave a remainder of 1.
17
103
121
The process of multiplication doesn't produce remainders.The process of division does.If you want to divide a 3-digit number by a one-digit numberand get a remainder of 8, try these:107 divided by 9116 divided by 9125 divided by 9134 divided by 9143 divided by 9..Add as many 9s to 107 as you want to, and then divide the result by 9.The remainder will always be 8.
27.2222
Regardless of the dividend (the number being divided), no divisor can produce a remainder equal to, or greater than, itself..... dividing by 4 cannot result in a remainder of 5, for example, Therefore the only single-digit number which can return a remainder of 8 is 9. 35 ÷ 9 = 3 and remainder 8
506
Any two digit number that fits the formula 3n+1 For example if n=4, 3x4+1=13 which fits your question.
I think you're wanting a number of two digits, one of which is 3, that when divided by 7 gives a quotient and a remainder of 1 and when that quotient is divided by 2 it gives a remainder of 1: Answer: 36 36 ÷ 7 = 5 r 1 5 ÷ 2 = 2 r 1 If you want the number to be such that if it is divided by 7 the remainder is 1 and if it is divided by 2 the remainder is 1, then: Answer: 43 43 ÷ 7 = 6 r 1 43 ÷ 2 = 21 r 1
The remainder is 8. To find the remainder when a number is divided by 9, add the digits together; if this sum has more than 1 digit, repeat until one digit remains. This digit is the remainder, unless it is 9 in which case the remainder is 0. examples: remainder 53 → 5 + 3 = 8 → remainder is 8 when 53 is divided by 9. remainder 126 → 1 + 2 + 6 = 9 → remainder is 0, ie no remainder, 126 is divisible by 9. remainder 258 → 2 + 5 + 8 = 15 → 1 + 5 = 6 → remainder is 6 when 258 is divided by 9.