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Factors & Remainders

Factors And Remainders

MODULES

HCF & LCM
Number of Factors
HCF with Remainders
LCM with Remainders
Operations on Remainders
Euler & Fermat
Special Types
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Factors & Remainders 1
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Factors & Remainders 2
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Factors & Remainders 3
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Factors & Remainders 4
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PRACTICE

Factors & Remainders : Level 1
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Factors & Remainders : Level 2
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Factors & Remainders : Level 3
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ALL MODULES

CAT 2025 Lesson : Factors & Remainders - HCF with Remainders

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3. Remainders

When a number, say nnn, is divided by a divisor (d)(d)(d), it leaves a quotient (q)(q)(q) and a remainder (r)(r)(r). So, nnn can be expressed as n\bm{n}n = dq\bm{dq}dq + r\bm{r}r

For example, when 373737 (n)(n)(n) is divided by 999 (d)(d)(d), we get a quotient of 444 (q)(q)(q) and remainder of 111 (r)(r)(r).

37=9×4+137 = 9 \times 4 + 137=9×4+1

Remainders are typically expressed as a non–negative value less than the divisor. ∴ 0≤r<d\bm{0 \leq r \lt d}0≤r<d

3.1 Adding/Subtracting Divisors

Property: If a multiple of the divisor is added to or subtracted from the number, the remainder does not change.

Let Kd\bm{Kd}Kd, be the multiple of the divisor that is added to the number, n\bm{n}n.

Rem(n+Kdd)=Rem(dq+r+Kdd)=Rem(d(q+k)d)+Rem(rd)\mathrm{Rem} \left(\dfrac{n + Kd}{d}\right) = \mathrm{Rem} \left(\dfrac{dq + r + Kd}{d}\right) = \mathrm{Rem} \left(\dfrac{d( q + k)}{d}\right) + \mathrm{Rem} \left(\dfrac{r}{d}\right)Rem(dn+Kd​)=Rem(ddq+r+Kd​)=Rem(dd(q+k)​)+Rem(dr​)

[As ddd(qqq +K+ K+K) is perfectly divisible by ddd, it's remainder is 000.]

=0+r=r= 0 + r = r=0+r=r

3.2 HCF concept with Remainders

HCF is applied in the following two types of questions on remainders.

Type 1: Finding the largest divisor that divides xxx, yyy and zzz to leave remainders aaa, bbb and ccc respectively.

When the numbers are subtracted from their respective remainders, they are perfectly divisible by the divisor.

∴ Largest Divisor === HCF[(xxx - aaa), (yyy - bbb), (zzz - ccc)]

Example 11

When each of 100100100, 125125125, 150150150 and 200200200 is divided by a natural number nnn, the remainders are 444, 555, 666 and 888 respectively. What is the largest possible value of nnn?

Solution

100100100 when divided by nnn leaves a remainder of 444.
This means n\bm{n}n perfectly divides 100−4=96100 - 4 = \bm{96}100−4=96.

∴n\bm{n}n perfectly divides (100−4100 - 4100−4), (125−5125 - 5125−5), (150−6150 - 6150−6) and (200−8200 - 8200−8) =96= \bm{96}=96, 120\bm{120}120, 144\bm{144}144 and 192\bm{192}192.

Largest number that can divide a set of numbers is it's HCF.
HCF(96,120,144,19296, 120, 144, 19296,120,144,192) =24= \bm{24}=24

Answer: 242424

Example 12

When each of 575757, 848484 and 102102102 is divided by xxx, the remainder is 333. What is the maximum possible value of xxx?

Solution

xxx perfectly divides (57−357 - 357−3), (84−384 - 384−3), (102−3102 - 3102−3) =54,81,99= 54, 81, 99=54,81,99

Largest value of x=x =x= HCF (545454, 818181, 999999) =9= 9=9

Alternatively

As the remainders are equal, the divisor should be a factor of the difference between numbers.

Difference between numbers === (84−5784 - 5784−57), (102−84102 - 84102−84) =27,18= 27, 18=27,18
HCF(272727, 181818) =9= 9=9

Answer: 999

Type 2: Finding the largest divisor that divides xxx, yyy and zzz to leave the same unknown remainder

As explained in the second solution of Example 121212 above, when the remainders are the same, then the largest possible divisor is the HCF of the difference between numbers.

Where x>y>zx \gt y \gt zx>y>z,
Largest Divisor === HCF[(x−yx - yx−y), (y−zy - zy−z)]

Example 13

What is the largest positive integer that divides 888888, 393939, 102102102 , and 606060, and leaves the same remainder?

Solution

Difference of numbers = (60−3960 - 3960−39), (88−6088 - 6088−60), (102−88102 - 88102−88) =21,28,14= 21, 28, 14=21,28,14

HCF(212121, 282828, 141414) =7= 7=7

Answer: 777

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