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)]`