## Remainders - LCM & HCF application

The basic concept of LCM & HCF is clubbed with the concept of remainders is required for solving certain types of questions.

## LCM + Remainders

### Type I : Common remainder

Any number which when divided by `x`, `y` or `z`, leaving the same remainder `r` in each of the divisions, is of the form `k( "LCM of x, y and z") + r`, where `k` is any whole number. When `k = 0`, we will get the least such value but this is the remainder itself and is less than the divisor.

if `k = 0; 35k + 4 = 4 -> "not prime"`

if `k = 1; 35 k + 4 = 39 -> "not prime"`

if `k = 2; 35 k + 4 = 74 -> "not prime"`

if `k = 3; 35k + 4 = 109 -> "prime"`

Therefore the required least prime = 109.

Largest three digit multiple of 99 is 990.`:. N = 990 + r`

Largest possible value for `r = 8`, because the least divisor is 9 and the remainder should be lesser than divisor.

`:. N = 990 + 8 = 998`

### Type II: Common Negative Remainders

When a number `N` divided by `x`, `y`, or `z`, it leaves the same negative remainder '-r', then the general form of such number `N = k ("LCM of x, y and z") - r`, where `k` is any natural number.

Divisors (d) | Positive remainders (r) | Negative remainders (r - d) |
---|---|---|

8 | 6 | `6 - 8 = -2` |

10 | 8 | `8 - 19 = -2` |

Required number `N = k("LCM of 25 and 30") - 5 = 150k - 5`

When `k = 6 -> N = 900 - 5 = 895`

`:.`The largest such three digit number = 895.

### Type III: Distinct remainders for different divisors.

Therefore `N = 7a + 2`, where 'a' is the quotient and 'a' is a natural number.

`"Rem" (N/11) = 7`

Therefore `N = 11b + 7`, where 'b' is the quotient and 'b' is a natural number.

`7a + 2 = 11b + 7`

`7a = 11b + 7 - 2`

`7a = 11b + 5`

LHS of the equation is a multiple of 7 for any natural number value for 'a'. so the RHS also should be a multiple of 7 for some suitable natural number values for 'b'.

When `b = 4 -> "RHS" = 11b + 5 = 11(4) + 5 = 49` , it is multiple of 7.

Hence 4 is the first such suitable value for 'b'.

So `N = 11b + 7 = 11(4) + 7 = 51`

The required least number = 51

## HCF + Remainders

### Type I: Distinct remainder for different dividends

The largest divisor 'd' with which the numbers `X`, `Y` and `Z` are divided leaves remainders `a`, `b` and `c` respectively is the HCF of `(X -a)`, `(Y - b)` and `(Z - c)`.

`= "HCF" [115, 161, 184]` = 23.

### Type II: Common remainder

The largest divisor which is giving the same remainder when it divides X, Y and Z is; `"HCF"[(X - Y),(Y - Z),(Z - X)]` OR `"HCF"[(X - Y),(Y - Z)]`