Remainders

Rem [2^1 /7] = 2

Rem [2^2 /7] = 4

Rem [2^3 /7] = 1

Rem [2^4 /7] = 2

Rem [2^5 /7] = 4

Rem [2^6 /7] = 1

So, the cycle/pattern is 2,4,1; the cyclicity is 3. Now, Rem [Power/Cyclicity] = Rem [32 /3] = 2. The answer is the 2^nd value in the list (i.e. 2,4,1), which is 4.

Note:

aⁿ when divided by b, will always give remainders which will have a pattern and will move in cycles of k such that k is less than or equal to b. Applying cyclicity to find remainders:

• Find out the cycle of remainders when aⁿis divided by b and make a list of those values.
• Find out the cyclicity, k
• Find out the remainder when the power is divided by the cyclicity i.e. Rem[n/k] = r
• The answer would be the r^thvalue in the list; if r = 0, it would be the last value in the list.
• While trying to find the cycle/pattern of remainders when aⁿ is divided by b, just multiply the previous remainder with ‘a’ to get the next value.

Rem [3^1 /11] = 3

Rem [3^2 /11] = 9

Rem [3^3 /11] = 5

Rem [3^4 /11] = 4

Rem [3^5 /11] = 1

Rem [3^6 /11] = 3

So, the cycle/pattern is 3,9,5,4,1; the cyclicity is 5. Now, Rem [Power/Cyclicity] = Rem [4567 /5] = 2. The answer is the 2^nd value in the list (i.e. 3,9,5,4,1), which is 9.

Note:

aⁿ when divided by b, will always give remainders which will have a pattern and will move in cycles of k such that k is less than or equal to b. Applying cyclicity to find remainders:

• Find out the cycle of remainders when aⁿis divided by b and make a list of those values.
• Find out the cyclicity, kFind out the remainder when the power is divided by the cyclicity i.e. Rem[n/k] = r
• The answer would be the r^thvalue in the list; if r = 0, it would be the last value in the list.
• While trying to find the cycle/pattern of remainders when aⁿ is divided by b, just multiply the previous remainder with ‘a’ to get the next value.

The remainder obtained by dividing a dividend by a divisor can replace that dividend for subsequent division. Hence, 4444 & 5555 in the given division can be replaced by 6 & 4 respectively, as these are the respective remainders obtained when 4444 and 5555 are divided by 7. Therefore, (4444⁵⁵⁵⁵ + 5555²²²²)/7 can be replaced by (6⁵⁵⁵⁵+4²²²²)/7.

Now, in 6⁵⁵⁵⁵/7,

Rem [6^1 /7] = 6

Rem [6^2 /7] = 1

Rem [6^3 /7] = 6

Rem [3^4 /7] = 1

Rem [3^5 /7] = 6

So, when the index of 6 is odd, the remainder is 6, and when the index is even, the remainder is 1. In this case, the index (5555) is odd and hence the remainder is 6.

Similarly, in 4²²²²/7,

Rem [4^1 /7] = 4

Rem [4^2 /7] = 2

Rem [4^3 /7] = 1

Rem [4^4 /7] = 4

So, the cycle/pattern is 4,2,1; the cyclicity is 3. Now, Rem [Power/Cyclicity] = Rem [2222/3] = 2. The answer is the 2^nd value in the list (i.e. 4,2,1), which is 2.

Therefore, (6⁵⁵⁵⁵+4²²²²)/7 is the same as (6+2)/7 which gives 1 as the remainder.

Note:

aⁿ when divided by b, will always give remainders which will have a pattern and will move in cycles of k such that k is less than or equal to b. Applying cyclicity to find remainders:

• Find out the cycle of remainders when aⁿis divided by b and make a list of those values.
• Find out the cyclicity, k
• Find out the remainder when the power is divided by the cyclicity i.e. Rem[n/k] = r
• The answer would be the r^thvalue in the list; if r = 0, it would be the last value in the list.

While trying to find the cycle/pattern of remainders when aⁿ is divided by b, just multiply the previous remainder with ‘a’ to get the next value.

2²⁸/13 = (2⁷)⁴/13

The remainder obtained by dividing a dividend by a divisor can replace that dividend for subsequent division. Hence,

(2⁷)⁴/13 = 11⁴/13 (remainder of 2⁷/13 is 11; so, replace 2⁷ with 11)

11⁴/13= (11²)²/13 = 4²/13 (remainder of 11²/13 is 4; so, replace 11² by 4). Now, 4²/13 = 16/13, which yields a remainder of 3.

When 561¹ is divided by 7, the remainder is 1

When 562¹ is divided by 7, the remainder is 2

When 561² is divided by 7, the remainder is 1² = 1 (by replacing the dividend 561 with the remainder of 561/7, which gives 1 as the remainder)

Apply the same concept as above,

When 562² is divided by 7, the remainder is 2² = 4

When 561³ is divided by 7, the remainder is 1³ = 1

When 562³ is divided by 7, the remainder is 2³ = 8, which when divided by 7 gives 1 as the actual remainder.

Therefore, for x = 3, the remainder of 561^x/7 and 562^x/7 will be the same.

Note:

The remainders will be the same for x = 3k where k is a natural number ≥ 1.

Rem [2^1 /9] = 2

Rem [2^2 /9] = 4

Rem [2^3 /9] = 8

Rem [2^4 /9] = 7

Rem [2^5 /9] = 5

Rem [2^6 /9] = 1

Rem [2^7 /9] = 2

So, the cycle/pattern is 2,4,8,7,5,1; the cyclicity is 6. Now, Rem [Power/Cyclicity] = Rem [553/6] = 1. The answer is the 1^st value in the list (i.e., 2,4,8,7,5,1), which is 2.

Note:

aⁿ when divided by b, will always give remainders which will have a pattern and will move in cycles of k such that k is less than or equal to b. Applying cyclicity to find remainders:

• Find out the cycle of remainders when aⁿis divided by b and make a list of those values.
• Find out the cyclicity, k
• Find out the remainder when the power is divided by the cyclicity i.e. Rem[n/k] = r
• The answer would be the r^thvalue in the list; if r = 0, it would be the last value in the list.

While trying to find the cycle/pattern of remainders when aⁿ is divided by b, just multiply the previous remainder with ‘a’ to get the next value.

Rem [2^1 /7] = 2

Rem [2^2 /7] = 4

Rem [2^3 /7] = 1

Rem [2^4 /7] = 2

Rem [2^5 /7] = 4

Rem [2^6 /7] = 1

So, the cycle/pattern is 2,4,1; the cyclicity is 3. Now, Rem [Power/Cyclicity] = Rem [9874/3] = 1. The answer is the 1^st value in the list (i.e. 2,4,1), which is 2.

Note:

aⁿ when divided by b, will always give remainders which will have a pattern and will move in cycles of k such that k is less than or equal to b. Applying cyclicity to find remainders:

• Find out the cycle of remainders when aⁿis divided by b and make a list of those values.
• Find out the cyclicity, k
• Find out the remainder when the power is divided by the cyclicity i.e. Rem[n/k] = r
• The answer would be the r^thvalue in the list; if r = 0, it would be the last value in the list.

While trying to find the cycle/pattern of remainders when aⁿ is divided by b, just multiply the previous remainder with ‘a’ to get the next value.

When a number is successively divided by two divisors d₁ and d₂ and two remainders r₁ and r₂ are obtained respectively, the remainder that will be obtained by the product of d₁ and d₂ is given by the relation d₁r₂ + r₁, where d₁ and d₂ are in ascending order respectively and r₁ and r₂ are their respective remainders when they divide the number. The number can be written in the form (d₁r₂ + r₁) * k(d₁d₂).

 

Here, divisors are 5 and 8; respective remainders are 2 and 3. Thus, d₁ = 5, d₂ = 8, r₁ = 2, r₂ = 3

-> Number is of the from 40k+17; least when k = 0

 

Therefore, the number is 17.

When a number is successively divided by three divisors d₁, d₂ and d₃, and three remainders r_1, r₂ and r₃ are obtained respectively, then applying the concept of successive division, the number can be written in the form:

[((d₂*r₃)+r₂)*d₁)+r₁]+ (d₁d₂d₃)k

Here, divisors are 5, 7 and 8; respective remainders are 2, 4 and 5. Therefore the number can be written in the form:

(5*7*8)k + [((7*5)+4)*5)+2] = 280k+197

Put k = 0

-> 280k+197 = 197

Therefore, the smallest such number is 197.

Divisors: 4, 5 and 7

Remainders: 2, 1 and 4

-> Number is of the form: 140k+86 (applying successive division for three divisors, as given in the note below)

-> Remainder is 86 when divided by 140, because 140k is divisible by 140 leaving no remainder, and the remaining portion, which is 86, is less than 140 and becomes the remainder.

Note:

When a number is successively divided by three divisors d₁, d₂ and d₃, and three remainders r_1, r₂ and r₃ are obtained respectively, then applying the concept of successive division, the number can be written in the form:

[((d₂*r₃)+r₂)*d₁)+r₁]+ (d₁d₂d₃)k