What Am I?

LCM of 1,2,3,4,5,6,7,8,9,10 is 2520. So we have to add or subtract 1 so that we will get the remainders mention in question. But if we add 1, conditions are not satisfied, so we must subtract 1. Hence ans is 2519.

call that positive number is x

x\2 remainder 1 => (x +1)\2 no remainder

x\3 remainder 2 => (x+1)\3 no remander

x\4 remainder 3 => (x+1)\4 no remainder

....

x\10 remainder 9 => (x+1)\10 no remainder

x+1 can divide 2,3,4,5,6,7,8,9,10

LCM of 2,3,4,5,6,7,8,9,10 is 2520

=> x+1 = 2520 => x = 2520 - 1 = 2519

divide by a number leaves a remainder 1 less from itself..

So, LCM of all numbers up until 10 is 2520.

2520-1=2519 is the required solution..

1st method

$(2a-1)=(3b-1)=(4c-1)=(5d-1)=(6e-1)$ $=(7f-1)=(8g-1)=(9h-1)=(10i-1)$ $=> 2a = 3b = 4c = 5d = 6e = 7f = 8g = 9h = 10i$ so, 2a is divisible by 3, 4, 5, 6, 7, 8, 9, 10 $$ hence least value of 2a is LCM of 3, 4, 5, 6, 7, 8, 9, 10 $=> 2a = 2520$ $=> 2a-1=2519$ $$ 2nd method almost similar $$ since 1 remains when divided by 2 => odd number $$ since 9 remains when divided by 10 => last digit is 9 $$ to get last digit as 9 $$ pattern is 90a-1, when divided by 9 and a = 1, 2, 3, 4, ........ $$ similarly $$ for 8, it is 40b-1 $$ for 7, it is 70c-1 $$ for 3 & 6 it follows 30d-1 $$ for 2 & 5 it follows 10e-1 $$ for 4 it shows 20f-1 $$ $$

to be the same number, $10e-1 = 20f-1 = 30d-1 = 40b-1 = 70c-1 = 90a-1$ $=> e=2f=3d=4b=7c=9a$ so, e is divisible by 2, 3, 4, 7 & 9 $$ and least value of e is LCM of 2, 3, 4, 7 and 9 $e = 252$ $=> 10\times252-1 = 2519$

7, 8, 9, 5 are co-prime integers, the result is: $5 \times 7 \times 8 \times 9 - 1= \boxed{2519}$

Take the LCM of 2 ,3 , 4 , 5 , 6 , 7 , 8 , 9 ,10

That is 2520

the required number is 2520 -1 =2519

No. ÷ 2 remainder 1 ... No. ÷ 10 remainder 9 here in each case (divisor – remainder) = 1

so least number = LCM of (2, 3, ..., 10) – 1

= 2520 – 1 = 2519

The unique prime factors of the 9 divisors collectively are 2, 3, 2, 5, 7, 2, 3 as derived below:

"2" = 2; "3" = 3; "4" = 2 (as one of the other 2's is already spoken for); "5" = 5; "6" = (null set; both the 2 and the 3 are already spoken for); "7" = 7; "8" = 2 (one new 2, the other two were accounted for above); "9" = 3 (one new 3); "10" -- (null set).

The product of these prime factors is 2520 which is also the LCM of all of the divisors combined (which is why we performed the prime factorization in the first place).

If we subtract 1 from the resulting 2,520, then the resulting dividend of 2,519 when divided by any of the 9 divisors should yield a quotient with a remainder of 1 less than the divisor (e.g., 3 for 4, 4 for 5, etc.), and that satisfies the initial conditions of the problem.

Can -1 be the solution...??????

Find the LCM of the numbers and subtract 1. Therefore 2519

I used MS excel vba code:

k = 1 For i = 1 To 10000 If (i Mod 2) = 1 And (i Mod 2) = 1 And (i Mod 3) = 2 And (i Mod 4) = 3 And (i Mod 5) = 4 And (i Mod 6) = 5 And (i Mod 7) = 6 And (i Mod 8) = 7 And (i Mod 9) = 8 And (i Mod 10) = 9 Then Cells(1 + k, 1) = i k = k + 1 End If Next i

Clearly, $n+1$ will be divisible by any integer $a$ such that $0<a<10$ . The smallest number that satisfies this condition is the $LCM$ of all the possible values of $a$ . This is equal to $2520$ . Thus $n=\boxed{2519}$

took a long time... take L.C.M of 1 2 3 4 5 6 7 8 9 10 and subtract 1

LCM(1,2,3,4,5,6,7,8,9,10) - 1

as the diff b/w the number and remainder is always same...(2-1=1,3-2=1,4-3=1.......10-9=1). so, we have to find LCM(2,3,4,5,6,7,8,9,10)=2520. we have to subtract that difference value which is 1 that we got above... therefore answer is 2520-1=2519

As stated N + 1 is divisible by 2, 3, 4, 5, 6, 7, 8, 9 & 10 so clearly one solution would be 10! (ie 10 9 8 7 6 5 4 3 2) but this is not the lowest possible. Follow the logic below...

N + 1 must be a multiple of 2

N + 1 must be a multiple of 4 but if it is a multiple of 2 & 4 it is necessarily a multiple of 8

N + 1 must be a multiple of 5 but if it is a multiple of 2 & 5 it is necessarily a multiple of 10

N + 1 must be a multiple of 6 but if it is a multiple of 2 & 9 (18) it is necessarily a multiple of 6

N + 1 must be a multiple of 7

N + 1 must be a multiple of 9

Hence 2 x 4 x 5 x 7 x 9 = 2520 is a multiple of 2, 3, 4, 5, 6, 7, 8, 9 & 10

Since the number leaves a remainder -1 when divided by 2,3,4...10 . It must leave a remainder -1 when divided by the lcm of 2,3,4,...10 .

To calculate the lcm simply take max power of prime numbers upto 10.

i.e. 8 * 9 * 5 * 7 = 2520

Hence, the answer is 2519.

you can solve it by finding the Least Common Multiple (LCM) of 2,3,4,5,6,7,8,9,10 and that is 2520, based on the information, the number always have a remainder of 1 if divided by 2,3,4,5,6,7,8,9,10 and that means the number can be 2520+1 or 2520-1, because the expression of each number is actually 2mod3,3mod4,4mod5,etc...9mod10, we must subtract the number by 1, hence the answer is 2520-1 = 2519

I think it's the LCM of 1,2,3...,10, which is 2520 (since it asks for the smallest positive integer), to which you apply the general algorithm you deduce from the hypothesis, which is n-1. So the final result would be 2520-1=2519

LCM OF 2,3,4,5,6,7,8,9,10=2520. on subtracting 1 from LCM will give the required number which gives 2519.

LCM of 2,3,4,5,6,7,8,9,10 is 2520 so this no. is divisible by all of them according to question let the no. be m than m+1 is divisible by 2,3,4,5,6,7,8,9,10 for it to be smallest it must be the LCM of the no.s respectively so m+1=2520 and m=2519

$In\quad terms\quad of\quad modulo,\quad there\quad are\quad 10\quad simultaneous\quad congruencies:\quad (x\quad is\quad the\quad desired\quad number)\\ x\equiv 9\left( mod10 \right) \\ x\equiv 8\left( mod9 \right) \\ and\quad so\quad on\quad until\\ x\equiv 1\left( mod2 \right) \\ If\quad we\quad add\quad 1\quad to\quad each\quad congruency,\quad then\quad the\quad remainder\quad would\quad be\quad 0.\quad Thus\quad we\quad let\quad x\quad have\quad a\quad factor\quad of\quad 2,3,another2,5,7,a\quad third\quad 2\quad and\quad another\quad 3.\quad Now\quad we\quad multiply\quad by\quad the\quad aformentioned\quad factors\quad and\quad subtract\quad 1\quad to\quad obtain\quad 2519$ .

Find out the L.C.M of numbers 2-10. It is 2520. Observe that the remainder is always one less than the divisor. Therefore subtract 1 from 2520. The answer is 2519.

simple lcm of all minus 1 2519

Let $x$ be the number sought. We then can proceed with the following system of modular equations.

$x == 9 (mod 10)$ -> $x = 10k + 9$

$10k + 9 == 8 (mod 9)$ -> $k == 8 (mod 9)$

$k = 9l + 8; x = 90l + 89$

$90l + 89 == 7 (mod 8)$ -> $2l + 1 == 7 (mod 8)$ -> $l == 3 (mod 8)$

$l = 8m + 3; x = 720m + 359$

$720m + 359 == 6 (mod 7)$ -> $6m + 3 == 0 (mod 7)$ -> $2m + 1 == 0 (mod 7)$ -> $2m + 7 == 6 (mod 7)$ -> $2m == 6 (mod 7)$ -> $m == 3 (mod 7)$

$m = 7n + 3; x = 5040n + 2519$

Notice that the remaining divisors (Numbers from 2 to 6) all divide 5040, so the number we're left to work with is 2519. Notice that it fulfills the conditions of the problem, and hence must be the solution.