**Shortcut for LCM **

__Type 1:__

**Remainder is same when divided by distinct divisors**

If the Number N leaves the same remainder R when divided by the numbers a, b, c, then the number N = (LCM of (a, b, c) p +R, where p is any whole number.

__Type 2:__

**Difference between the divisors and remainders is same**

If the number N is divided by the numbers a, b, c and N gives three different remainders p, q, r respectively such that the difference between the divisors and remainders is same i.e.(a-p)=(b-q)=(c-r)=Z, then the number N must be in the form of {(LCM of a, b, c) p – Z} in each case, where p is any whole number.

__Type 3:__

**Neither remainder is same nor is difference between the divisors and remainders same.**

When the number N is divided by the two numbers m and n and they leave remainders p, q respectively. In this case we identify the smallest number (say A) that satisfies given conditions (the smallest number which gives remainder p, q when divided by m, n and this number is identified by equating the number and using the fact that quotient is whole number). Once we have done that, the number must be in the form of {(LCM of a, b)p + A} (A is the smallest number)and p is any whole number.

**Let’s solve few examples based on above shortcuts:**

**Example 1: **Find the largest 4 digit number, which gives 4 as a remainder when divided by 17 and 19?

** ****Solution:**

Let N be such number which satisfies given conditions.

Since N gives the same remainder 4 in each case. Hence N must be of the form

N = (LCM of 17, 19) p + 4where p is any whole number

Since LCM of 17 and 19 is 323 so the number must be in the form

N=323 p + 4

The largest value of “p” possible so that the N remains 4-digit number is 30.

Therefore, 323p = 323 X 30=9690.

Hence largest 4-digit number, which gives 4 as a remainder when divided by 17 and 19

is 9694.

**Example 2**: Find the largest 4-digit number, which gives the remainder 7 and 13 when divided by 11 and 17?

**Solution: **

Let N be such number satisfying given conditions

Here remainders are not same but the difference between the remainders and the divisors is same i.e. 11-7 =17-13=4

Hence N must be of the form

N = (LCM of 11, 17) p – 4where p is any whole number

Since LCM of 11 and 17 is 187so the number must be in the form

N=187 p – 4.

The largest value of “p” possible so that the N remains 4-digit number is 53.

Therefore, 187p= 187 x 53=9911.

Hence the largest 4-digit number, which gives remainders 7 and 13 when divided by 11 and 17, is 9907.

**Example 3**: Find the largest 3-digit number, which gives the remainder 1 and 2 when divided by 3 and 5?

**Solution: **

Let N be such number satisfying given conditions

i.e. N = 3a + 1 where a is any whole number

&N = 5b + 2 where b is any whole number.

Where a & b are whole numbers.

Equating

3a + 1=5b + 2

b= (3a-1)/5

Since b is whole number. So, 5 should divide numerator implies a=2.

Thus smallest number satisfying both given conditions is 7.

So the number will be in the form of (LCM of 3,5)k + 7.

Now we have to find the largest 3-digit number N in the form of 15k+7

The largest value of “k” possible so that the N remains 3-digit number is 53.

Hence the largest 3-digit number, which gives the remainder 1 and 2 when divided by 3 and 5 is 15 x 66 + 7 = 997

Hey @murali,

The smallest possible number which satisfies the condition for b is a whole number will be a=2.

Now, putting the value a=2 in the equation, (3a-1)/5, we get b=1.

So, a = 2 and b = 1,

N=3a+1=3(2)+1=7

N=5b+2=5(1)+2=7

So, the smallest number satisfying both given conditions is 7.

I hope you got the answer

Since b is whole number. So, 5 should divide numerator implies a=2.

Thus smallest number satisfying both given conditions is 7.

can anyone u please explain how it was