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A function D(a, 10b + c) is defined as the remainder when the sum a0 + 09 Jun 2020, 07:13
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A function D(a, 10b + c) is defined as the remainder when the sum \(a^0 + a^1 + ... + a^{10b + c}\) is divided by c, where a, b and c are single-digit positive integers. What is the value of D(y, 10x + z) where x, y and z are single-digit positive integers such that x < y < z , x and z are perfect squares and the difference between the sum and the product of the prime factors of y is 1?

A. 0
B. 3
C. 6
D. 7
E. Cannot be determined


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D
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A function D(a, 10b + c) is defined as the remainder when the sum a0 + 09 Jun 2020, 07:58
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y must be equal to 6. [(3*2)-(3+2) =1]

since z is a perfect square and greater than y=6, z must be equal to 9.

x can be 1 or 4

Case 1- x= 1

\(6^n\) must be divisible by 9, for n>1

hence, you can write

\([6^0+6^1+6^2+........+6^{19}]\) = 6^0 + 6 + 9k = 9k+7

D(6, 19) = 7

Case 2- x= 4

Similarly we can write

\([6^0+6^1+6^2+........+6^{49}]\) = 1+6+9m = 9m+7

D(6, 49) = 7

Answer - D

Bunuel
A function D(a, 10b + c) is defined as the remainder when the sum \(a^0 + a^1 + ... + a^{10b + c}\) is divided by c, where a, b and c are single-digit positive integers. What is the value of D(y, 10x + z) where x, y and z are single-digit positive integers such that x < y < z , x and z are perfect squares and the difference between the sum and the product of the prime factors of y is 1?

A. 0
B. 3
C. 6
D. 7
E. Cannot be determined


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A function D(a, 10b + c) is defined as the remainder when the sum a0 + 09 Jun 2020, 08:30
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[quote="nick1816"]y must be equal to 6. [(3*2)-(3+2) =1]

since z is a perfect square and greater than y=6, z must be equal to 9.

x can be 1 or 4

Case 1- x= 1

\([6^0+6^1+6^2+........+6^{19}]\) mod 9 = \([6^0+6]\) mod 9 = 7 mod 9

Case 2- x= 4

\([6^0+6^1+6^2+........+6^{49}]\) mod 9 = \([6^0+6]\) mod 9 = 7 mod 9

D

would you please elaborate on the procedure of finding mod?
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A function D(a, 10b + c) is defined as the remainder when the sum a0 + Updated on: 09 Jun 2020, 10:39
Originally posted by yashikaaggarwal on 09 Jun 2020, 09:20.
Last edited by yashikaaggarwal on 09 Jun 2020, 10:39, edited 1 time in total.
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X Y and Z are single digit number
X and Z = single digit perfect square = 1,4,9
Y = single digit number whose prime factor sum and product difference is 1 =
Let Y = {9,8,6,4}

Putting Y = 9 (prime factors =3^2 = single prime factor)
(not Satisfying)

Putting Y = 8 (prime factors =2^3 = single prime factor)
(not Satisfying)

Putting Y = 6
Sum of prime factor = 2+3 = 5
Product of prime factor = 3*2 = 6
Difference = 6-5 = 1 (Satisfying)

Putting Y = 4 (prime factors =2^2 = single prime factor)
(not Satisfying)

Therefore Y = 6
Z = 9
And X = 1 or 4

D(Y,10X+Z) = Y^0+Y^1+Y^2+..........Y^19 or Y^49 divided by Z

6^0+6^1+6^2+..........6^19 or 6^49 divided by 9

1 mod 9 +(6^1+6^2+..........6^19 or 6^49) mod 9
1 mod 9 = 1
6^even mod 9 = 0
6^odd mod 9 = 6
6^19 or 6^49 both are odd
Therefore 6^19 or 6^49 mod 9 = 6
(1+6) mod 9 = 7 mod 9 = 7
Answer is D

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A function D(a, 10b + c) is defined as the remainder when the sum a0 + 09 Jun 2020, 10:25
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minustark

\(6^n\) must be divisible by 9, for n>1

hence, you can write

\([6^0+6^1+6^2+........+6^{19}]\) = 6^0 + 6 + 9k = 9k+7

a = x mod n
a , when divided by n, leaves x as remainder

(a+b) mod n = a mod n +b mod n

(a*b) mod n = a mod n * b mod n

You don't need it for GMAT tho



yashikaaggarwal

factors of 9 = 1,3,9
only 3 is a prime factor

You shouldn't consider 2 3's as prime factor of 9. [same with other numbers]
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A function D(a, 10b + c) is defined as the remainder when the sum a0 + 09 Jun 2020, 10:35
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nick1816
minustark

\(6^n\) must be divisible by 9, for n>1

hence, you can write

\([6^0+6^1+6^2+........+6^{19}]\) = 6^0 + 6 + 9k = 9k+7

a = x mod n
a , when divided by n, leaves x as remainder

(a+b) mod n = a mod n +b mod n

(a*b) mod n = a mod n * b mod n

You don't need it for GMAT tho



yashikaaggarwal

factors of 9 = 1,3,9
only 3 is a prime factor

You shouldn't consider 2 3's as prime factor of 9. [same with other numbers]
Show more
Well I just thought both are prime factors so.... Never mind. Will correct. Thanks though.

Posted from my mobile device
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A function D(a, 10b + c) is defined as the remainder when the sum a0 + 09 Jun 2020, 18:49
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Bunuel
A function D(a, 10b + c) is defined as the remainder when the sum \(a^0 + a^1 + ... + a^{10b + c}\) is divided by c, where a, b and c are single-digit positive integers. What is the value of D(y, 10x + z) where x, y and z are single-digit positive integers such that x < y < z , x and z are perfect squares and the difference between the sum and the product of the prime factors of y is 1?

A. 0
B. 3
C. 6
D. 7
E. Cannot be determined
Show more
Ans : D

x, y and z are single-digit positive integers
such that x < y < z ,
x and z are perfect squares
and the difference between the sum and the product of the prime factors of y is 1


X---------------------Y------------------Z

1--------------------Y-------------------4 (Now Y can be 2 or 3 .. but in both the case difference between the sum and the product of the prime factors of y is not 1 . So this combination not possible )




4--------------------Y-------------------9 (Y can 5 , 6 ,7 , 8 .. Only 6 satisfy the condition .. 2*3 - (2+3) = 1 .)
So Y has to be 6 .



9--------------------Y------------------[Not possible]




Now D(y, 10x + z) = \((6^0 + 6^1 + 6^2 ..........+6^{49} )/9\)
Out of all the term only \(6^0+6^1 =7\) won't be divisible by 9 . Hence the reminder will be 7 .

Answer D .
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A function D(a, 10b + c) is defined as the remainder when the sum a0 + 11 Jun 2020, 13:28
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Bunuel
A function D(a, 10b + c) is defined as the remainder when the sum \(a^0 + a^1 + ... + a^{10b + c}\) is divided by c, where a, b and c are single-digit positive integers. What is the value of D(y, 10x + z) where x, y and z are single-digit positive integers such that x < y < z , x and z are perfect squares and the difference between the sum and the product of the prime factors of y is 1?

A. 0
B. 3
C. 6
D. 7
E. Cannot be determined

Show more

Solution:

We need to determine the remainder when y^0 + y^1 + y^2 + … + y^(10x + z) is divided by z.

We are given that x, y, and z are single-digit positive integers. Since x and z are perfect squares, they are 1, 4 or 9. Since the the difference between the sum and the product of the prime factors of y is 1, y must be 6 (notice that the prime factors of 6 are 2 and 3, and (2 * 3) - (2 + 3) = 1). Since x < y < z, we have 2 cases: 1) x = 1, y = 6, and z = 9, and 2) x = 4, y = 6, and z = 9.

Case 1: x = 1, y = 6 and z = 9

We need to determine the remainder when 6^0 + 6^1 + 6^2 + … + 6^19 is divided by 9.

Notice that all the terms in the above sum are divisible by 9 starting from 6^2. Therefore, the only terms that are not divisible are 6^0 and 6^1. Since 6^0 + 6^1 = 1+ 6 = 7 and when 7 is divided by 9, the remainder is 7. Therefore, the remainder when 6^0 + 6^1 + … + 6^19 divided by 9 is 7.

Case 2: x = 4, y = 6 and z = 9

We need to determine the remainder when 6^0 + 6^1 + 6^2 + … + 6^49 is divided by 9.

As in case 1, all the terms in the above sum are divisible by 9 starting from 6^2. Therefore, the only terms that are not divisible are 6^0 and 6^1. Therefore, the remainder is also 7.

Answer: D
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A function D(a, 10b + c) is defined as the remainder when the sum a0 + 13 Jun 2020, 08:26
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yashikaaggarwal
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minustark

\(6^n\) must be divisible by 9, for n>1

hence, you can write

\([6^0+6^1+6^2+........+6^{19}]\) = 6^0 + 6 + 9k = 9k+7

a = x mod n
a , when divided by n, leaves x as remainder

(a+b) mod n = a mod n +b mod n

(a*b) mod n = a mod n * b mod n

You don't need it for GMAT tho



yashikaaggarwal

factors of 9 = 1,3,9
only 3 is a prime factor

You shouldn't consider 2 3's as prime factor of 9. [same with other numbers]
Well I just thought both are prime factors so.... Never mind. Will correct. Thanks though.

Posted from my mobile device
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Got it..thank you.
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A function D(a, 10b + c) is defined as the remainder when the sum a0 + 16 Jun 2026, 01:30
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A function D(a, 10b + c) is defined as the remainder when the sum \(a^0 + a^1 + ... + a^{10b + c}\) is divided by c, where a, b and c are single-digit positive integers.

What is the value of D(y, 10x + z) where x, y and z are single-digit positive integers such that x < y < z , x and z are perfect squares and the difference between the sum and the product of the prime factors of y is 1?

Single-digit perfect squares = {1, 4, 9}

Possible combinations
x=1;z=4
x=1;z=9
x=4;z=9

y = 1,2,3,5,7; Sum of prime factors = Product of prime factors ; Difference = 0
y = 4; Sum of prime factors = 2+2=4; Product of prime factors=2*2=4 ; Difference = 0
y = 6; Sum of prime factors = 2+3=5; Product of prime factors=2*3=6 ; Difference = 1
y = 8; Sum of prime factors = 2+2+2=6; Product of prime factors=2*2*2=8 ; Difference = 2
y = 9; Sum of prime factors = 3+3=6; Product of prime factors=3*3=9 ; Difference = 3

y = 6
Possible combinations since x<y<z
x=1; y=6; z=9; D(y, 10x + z) = D(6, 10*1 + 9) = D(6,19) = The remainder when the sum \(6^0 + 6^1 + ... + 6^{19}\) is divided by 9 = 1 + 6 + 0+...+0 = 7
x=4; y=6; z=9; D(y, 10x + z) = D(6, 10*4 + 9) = D(6,49) = The remainder when the sum \(6^0 + 6^1 + ... + 6^{49}\) is divided by 9 = 1 + 6 + 0+...+0 = 7

Let us understand the behavior of the remainder when 6^i is divided by 9
6^0 = 1 (mod 9)
6^1 = 6 (mod 9)
6^2 = 36 = 0 (mod 9)
6^i = 0 (mod 9) for all i>=2

The remainder when 6^0 + 6^1 + .... + 6^n is divided by 9= 1 + 6 = 7

IMO D
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A function D(a, 10b + c) is defined as the remainder when the sum a0 + 16 Jun 2026, 05:58
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A function D(a, 10b + c) is defined as the remainder when the sum \(a^0 + a^1 + ... + a^{10b + c}\) is divided by c, where a, b and c are single-digit positive integers. What is the value of D(y, 10x + z) where x, y and z are single-digit positive integers such that x < y < z , x and z are perfect squares and the difference between the sum and the product of the prime factors of y is 1?

A. 0
B. 3
C. 6
D. 7
E. Cannot be determined


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x, y and z are single-digit positive integers such that x < y < z ,

x, y and z have values between 1 - 9 such that x < y < z

x and z are perfect squares

There are only 3 perfect squares between 1 to 9 -> 1, 4, 9
So {x, z} = {1, 4} or {1, 9} or {4, 9}

and the difference between the sum and the product of the prime factors of y is 1

If y were a prime number, its prime factor would be only itself and hence sum and product of prime factors would be equal. So y cannot be 2/3/5/7.
y could be 4 or 6 or 8. It cannot be 9 because z > y.
Only for 6, do we see that sum of prime factors (2+3) and product of prime factors (2*3) differ by 1.
So y = 6.

Hence x, y, z can be 1, 6, 9 or 4, 6, 9. So y is 6 and z is 9 for sure.

D(y, 10x + z) is remainder when \(6^0 + 6^1 + 6^2 + ....6^{19} / 6^{49}\) is divided by 9.

Note that all 6^2, 6^3, 6^4 and so on will be divisible by 9 because 6 has 3 as a factor and 9 needs two 3s. All these terms will have at least two 6s which means they will have at least two 3s. So all these terms will be divisible by 9.

Hence we ignore all exponents of 6 except the first two: \(6^0 + 6^1 = 7\)

Remainder when (7 + 9a + 9b + ...) is divided by 9 will be 7.

Answer (D)
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