Topic - Number Theory

Solved ? - Yes

Source - ]]>

Topic - Number Theory

Solved ? - Yes

Source - ]]>

a = mq + r1

b = nq + r2

Then, (a + b) = (m + n)q + (r1 + r2)

Remainder when sum of two numbers (giving individual remainders r1 and r2 with a divisor) is divided by the same divisor is :

R = (r1 + r2) - divisor [if this is positive]

else

R = (r1 + r2)

Here R = (3 + 4) - m

R + m = 7 ]]>

4 only-multiples of 31

30 only-multiples of 5

1 multiple of both 31 and 5

35 such numbers

Hence,

Sum = 4 * 31 + 30 * 5 + 1 * 155 + (155 - 35) * 1 = 549

This is an example of how test setters combine two simple concepts to trap you.

]]>By remainder theorem,

Remainder when n^3 + 83 is divided by (n - 11) = 11^3 + 83 = 1414

Now,

(n - 11) * k = 1414

For largest n,

k = 1, (n - 11) = 1414

n = 1425 ]]>

I. (de)^2 = abbb

II. (ee)^2 = ccbb

III. A perfect square never ends with c,d and e.

Which of the following statements are sufficient to determine the arithmetic mean of a, b and c?

a) Any of the three statements

b) Any two of the three statements

c) Statement I and II are sufficient

d) Statement II and III are sufficient

e) All three statements are necessary

The only square of the form XX^2 = YYZZ is 88^2 = 7744

A square never ends with 2,3,7,8 or odd number of zeroes

Combine them, analyze the options and you ll find I and II are sufficient to answer what we need which is primarily values of a,b,c. ]]>

When we cancel out some common term from dividend and divisor, we need to multiply it back in the end.

24^1202 = (2^3 * 3)^1202 = 2^3606 * 3^1202

2^3605 * 3^1202 mod 3*241

Use Chinese remainder theorem now,

N = 2^3605 * 3^1202 N mod 3 = 0

E(241) = 240

N mod 241 = 2^3605 * 3^1202 mod 241 = (2^5 * (2^240)^15) * (3^2 * (3^240)^5) mod 241

[Taking out powers in multiple of Euler number, as they would leave remainder as 1 with 241.]

N mod 241 = 2^5 * 3^2 mod 241 = 47

3a = 241b + 47

a = (240b + 45)/3 + (b + 2)/3

a = 80b + 15 + (b + 2)/3

b = 1, a = 96

Remainder = 288

Final remainder = 288*2 = 576 [ Remember we needed to multiply the 2 canceled out earlier back]

Alternate approach

We know that for a prime p, N^(p – 1) mod p = 1 ; By extension of Euler theorem which is also called Fermat’s remainder theorem. That is to say, Cyclicity of obtaining remainder 1 is E (p).

Extending that logic, 1446 = 2 * 3 * 241

Cyclicity of 1446 = LCM (E (2), E (3), E (241)) = LCM (1, 2, 240) = 240

24^1202 mod 1446 = (24^240)^5 * 24^2 mod 1446 = 1 * 576 mod 1446 = 576

h^3. pqr = 12h

h^2. pqr = 12

h = 1 => pqr = 12 => 18 ordered sets

h = 2 => pqr = 3 => 3 ordered sets

Total : 21 ordered sets ]]>

N = 101,206,311,416,521,626

626 mod 35 = 31 ]]>

Multiply and divide by 2 * 4 * 6 * 8 *... * 998 * 1000

Number = 1001!/(2 * 4 * 6 * 8 * .. * 998 * 1000) = 1001!/2^500 * 500!

Exponent of 13 in this = (77 + 5) - (38 + 2) = 42

Approach 2 :

For power of 13, 13 * 1 to 13 * (7 * 11) all odd powers

So, (7 * 11 + 1)/2 = 39

For 13^2,

13 * (13 * 1), 13 * (13 * 3), ... up-to 13 * (13 * 5)

So additional three powers.

Total = 39 + 3 = 42

a) 24

b) 60

c) 64

d) Cannot Be Determined ]]>

15 can be broken into three factors as follows - (5, 3, 1) or (15, 1, 1).

Solving for p, q, r we get (4, 3, 2) and (8, 8, 1) as the triplets corresponding to the above set of factors.

Thus product of p, q and r can be either 24 or 64. ]]>

LCM = hab since LCM * HCF = ha * hb

So,

h(a + b + ab) = p

Case 1 : h != 1

Then, h = p => prime => Not possible

Since LCM >= Numbers >= HCF always

Case 2 : h = 1

a + b + ab = p

(a + 1)(b + 1) = (p + 1)

Number of pairs = [Factors of (p + 1)]/2 - 1 for the case when a becomes zero.

For the given range,

p = 11, 13, 17, 19, 23, 29

p = 11 => 12 = 2^2 * 3 => 3 * 2/2 - 1 = 2 pairs

p = 13 => 14 = 2 * 7 => 2 * 2/2 - 1 = 1 pair

p = 17 => 18 = 2 * 3^2 => 2 pairs

p = 19 => 20 = 2^2 * 5 => 2 pairs

p = 23 => 24 = 2^3 * 3 => 4 * 2/2 - 1 = 3 pairs

p = 29 => 30 = 2 * 3 * 5 => 2 * 2 * 2/2 - 1 = 3 pairs

So 2 such p = 23,29 ]]>