Number System Tutorial Part III: Factors, Multiples, Unit Digit and Last Two Digits of Exponents

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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.

Find the least prime number which leaves a remainder of 4, when it is divided by 5 or 7.?
Least number which leaves the remainder 4 when divided by 5 or 7 is: `k ("LCM of 5 and 7") + 4 = 35k + 4`
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.
Find the largest three digit number that leaves a common remainder when it is divided by 9 or 11.?
`N = k ("LCM of 9 and 11") + r = 99k + r`
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.

Find the smallest number which when divided by 8 or 10, leaves the remainders 6 and 8 respectively.?
Divisors (d)Positive remainders (r)Negative remainders (r - d)
8 6 `6 - 8 = -2`
10 8 `8 - 19 = -2`
The required number `N = "LCM"(8, 10) - 2 = 40 - 2 = 38`
Find the largest three digit number which leaves remainders 20 and 25 when dived by 25 and 30 respectively.?
Negative remainder in each case; `20 - 25 = -5` and `25 - 30 = -5`
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.

Find the smallest number which leaves remainders 2 and 7 when divided by 7 and 11 respectively.?
Let the number be 'N'; `"Rem" (N/7) = 2`
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)`.

Find the largest divisor which gives the remainders 5, 9 and 6 when it divides 120, 170 and 190 respectively.?
Required divisor = `"HCF" [ (120 - 5), ( 170 - 9), (190 - 6)]`
`= "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)]`

Find the largest divisor which divides 239, 380 and 521, leaving the same remainder in each division.?
Required divisor = `"HCF" [(380 - 239),(521 - 380)] = HCF [47,141] = 47`
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