Interesting Number Properties

Question

I will keep on adding basis my learning from various questions and post, new learning will be added in the form of new post

5 Rules Worth Memorizing
 
 Difference of squares ⇔ odd or multiple of 4. 
 Odd sum of two primes ⇒ one prime must be 2. 
 In factorials, cancel first; never expand. 
 AM ≥ GM with equality at x = y. 
 Equal numbers often create extrema:
• Maximum product for fixed sum
• Minimum sum for fixed product
• Minimum sum of squares for fixed sum.

More theory

AM-GM Thinking
Formula:
(x+y)/2 ≥ √(xy)
Equality only when:
x = y
 Most Important Learning:
 For a fixed sum:
Product is maximum when numbers are equal.
Example:
Sum = 20
10×10 = 100
9×11 = 99
8×12 = 96
Maximum occurs at:
10 and 10
Second Learning:
For a fixed product:
Sum is minimum when numbers are equal.
Example:
Product = 36
6+6 = 12
4+9 = 13
3+12 = 15
Minimum occurs at:
6 and 6
GMAT Trigger:
See:
• Fixed sum + maximize product
OR
• Fixed product + minimize sum
Immediately test:
x = y

Minimum/Maximum Strategy
Key Rule:
For a fixed sum:
x2 + y2
is minimized when:
x = y
Example:
x+y = 10
Case 1:
52 + 52
= 25 + 25
= 50
Case 2:
42 + 62
= 16 + 36
= 52
Case 3:
22 + 82
= 4 + 64
= 68
Notice:
The more unequal the numbers become, the larger the sum of squares becomes.
GMAT Example:
Given:
(x+y)2 > 200
Find minimum possible value of:
x2 + y2
Minimum occurs when:
x = y
So:
(2x)2 > 200
4x2 > 200
x2 > 50
Therefore:
x2 + y2 > 50 + 50 = 100
GMAT Trigger:
When asked to minimize:
x2+y2
a2+b2
sum of squares
Try making variables equal.

Difference of Squares
Formula:
N = a2 − b2 = (a+b)(a−b)
Key Rule:
A number can be written as a difference of squares if:
• It is odd
OR
• It is divisible by 4
Why?
(a+b) and (a−b) are either both odd or both even.
Odd × Odd = Odd
Even × Even = Multiple of 4
Example:
15 = 42 − 12 = (4+1)(4−1) = 5×3
20 = 62 − 42 = (6+4)(6−4) = 10×2
Counterexample:
6 cannot be written as a2 − b2
Reason:
6 = 4k+2
Numbers of form 4k+2 never work.

Prime Number Sums
If the sum of two primes is odd:
One of the primes must be 2.
Example:
19 = 2 + 17
Works because 17 is prime.
GMAT Trigger:
When asked:
"Can odd number N be written as sum of two primes?"
Check:
N − 2
If N−2 is prime, answer is Yes.
If not, answer is No.
Example:
45 − 2 = 43 (prime)
So:
45 = 2 + 43
Works.

Usernamevisible · Answer

The key idea:
When you add the same positive number b to both the numerator and denominator,
x/y → (x+b)/(y+b)
the new fraction moves closer to 1.
Why?
Because the numerator and denominator become more similar.
----------------------------------------------------------

Case 1: Fraction less than 1
2/5 = 0.4
Add 3 to both:
(2+3)/(5+3) = 5/8 = 0.625
The fraction got bigger and moved closer to 1.
Therefore:
If x/y < 1, then (x+b)/(y+b) > x/y
----------------------------------

Case 2: Fraction greater than 1
5/2 = 2.5
Add 3 to both:
(5+3)/(2+3) = 8/5 = 1.6
The fraction got smaller and moved closer to 1.
Therefore:
If x/y > 1, then (x+b)/(y+b) < x/y
----------------------------------

GMAT Shortcut:
If a fraction is less than 1, adding the same number to the numerator and denominator makes it larger.
If a fraction is greater than 1, adding the same number to the numerator and denominator makes it smaller.
In both cases, the fraction moves toward 1.

Key Rule
Start with x/y.
Add a to the numerator and b to the denominator:
(x + a)/(y + b)
The new fraction moves closer to a/b.
Decision Rule
If x/y < a/b, the fraction increases.
If x/y > a/b, the fraction decreases.
If x/y = a/b, the fraction stays the same.
Example 1
Start: 2/7
Add 3 to numerator and 5 to denominator.
Target fraction = 3/5.
Since 2/7 < 3/5, the new fraction is larger:
2/7 < 5/12
Example 2
Start: 11/12
Add 2 to numerator and 5 to denominator.
Target fraction = 2/5.
Since 11/12 > 2/5, the new fraction is smaller:
11/12 > 13/17
GMAT Shortcut
Compare the original fraction x/y with a/b.
If x/y < a/b → fraction goes up.
If x/y > a/b → fraction goes down.
If x/y = a/b → no change.
Think of (x + a)/(y + b) as a weighted average of x/y and a/b, so it always moves toward a/b.

Key rule:
(x+2)/(y+3) moves toward 2/3.
Need to know whether x/y is above or below 2/3.
(1) y > 20
No information about x. x/y could be above or below 2/3.
Insufficient.
(2) x < 5
No information about y. x/y could be above or below 2/3.
Insufficient.
Together:
x < 5 and y > 20
Largest possible fraction is 4/21.
Since 4/21 < 2/3, x/y < 2/3 always.
Moving toward 2/3 means the fraction increases.
Definite answer: Yes.
Answer = C.
Key rule:
(x+5)/(y+5) moves toward 1.
Need to know whether x/y is above or below 1.
(1) y = 5
No information about x.
x/y could be less than or greater than 1.
Insufficient.
(2) x > y
Since x/y > 1, the fraction is above 1.
Moving toward 1 makes the fraction smaller.
Definite answer: No.
Sufficient.
Answer = B.

Usernamevisible · Answer

Percentile
If a person's score is in the xth percentile, it means they scored higher than x% of the group.
GMAT shortcut:
xth percentile out of n people → outscored x% × n people.
---------------------------------------------------------

Example
Lena's grade was in the 80th percentile out of 120 students.
In another class of 200 students, there were 24 grades higher than Lena's.
Nobody had the same grade as Lena.
Find Lena's percentile in the combined group.
---------------------------------------------

Step 1: Students Lena outscored in first class
80th percentile of 120
120 × 0.80 = 96
Lena outscored 96 students.
---------------------------

Step 2: Students Lena outscored in second class
200 − 24 = 176
Lena outscored 176 students.
----------------------------

Step 3: Combine
Total students outscored
96 + 176 = 272
Total students
120 + 200 = 320
---------------

Step 4: Find percentile
272/320 = 0.85 = 85%
Lena is in the 85th percentile.
-------------------------------

GMAT Takeaway
Percentile → convert directly into "number of people outscored."
Then:
Combined Percentile = (Total People Outscored) / (Total People) × 100%

Usernamevisible · Answer

Common Factors Rule (Advanced)
Key Idea:
Any common factor of two numbers must divide their difference.
Example 1
476 and 478
Difference:
478 − 476 = 2
Factors of 2:
1, 2
Therefore the only possible common factors are:
1, 2
Check:
476 = 2 × 238
478 = 2 × 239
GCD = 2
-------

Example 2
476 and 484
Difference:
484 − 476 = 8
Factors of 8:
1, 2, 4, 8
Therefore any common factor must be one of:
1, 2, 4, 8
Nothing else is possible.
Check:
476 = 4 × 119
484 = 4 × 121
GCD = 4
-------

Important GMAT Shortcut
If two numbers differ by a small number, their GCD must be a factor of that difference.
Example
Numbers differ by 6.
Possible GCD values:
1, 2, 3, 6
No other GCD is possible.
-------------------------

-------

---

Very Common GMAT Use
Question:
The GCD of two integers is 15. Which of the following could be their difference?
Rule:
GCD must divide the difference.
Since GCD = 15,
Difference must be a multiple of 15:
15, 30, 45, 60, 75, ...
Any non-multiple of 15 is impossible.
-------------------------------------

Reverse Use
Numbers:
203 and 218
Difference:
15
Since GCD must divide 15,
Possible GCD values:
1, 3, 5, 15
No other value is possible.
---------------------------

Credit to Bunuel    
   Question:  Given that x is a positive integer, what is the greatest common divisor (GCD) of the two positive integers, (x+m) and (x-m)?

Statement 1:  
  m2–10m+16=0m2–10m+16=0

Statement 2:  
  x+26x+26  
  is a prime number.

Solution:

The two given positive integers are (x + m) and (x – m). x is a positive integer so m must be an integer too. Whether m is positive or negative, we don’t know.

To know the GCD of two numbers, we need to know their common divisors. As of now, we have no idea about their common divisors, but we know that the difference between the two numbers is 2m. Their common factors must be factors of 2m.

Let’s look at the two statements:

Statement 1:   
  m2–10m+16=0m2–10m+16=0

We know that the quadratic will give us two values for m so we will not be able to find a unique value for m. But let’s solve it in case we get some other clues from it.

m2–10m+16=0m2–10m+16=0

m2–2m–8m+16=0m2–2m–8m+16=0

m(m–2)–8(m–2)=0m(m–2)–8(m–2)=0

(m–2)∗(m–8)=0(m–2)∗(m–8)=0

m is either 2 or 8. So 2m is either 4 or 16.

The factors of 2m will be 1, 2 and 4 and additionally, 8 and 16 (if 2m is 16). We have no idea whether x+m and x-m will have these factors so this statement alone is not sufficient.

Statement 2:   
  x+26x+26  
  is a prime number.

What does it tell us about x? Other than 2, all prime numbers are odd numbers. Since x is a positive integer, x+26 cannot be 2. It must be a prime number greater than 2 and hence, must be odd. But 26 is even. So x must be an odd integer (Odd + Even = Odd). But we have no information about m so this statement alone is not sufficient.

Using both statements together, since x is an odd integer and m is definitely even (either 2 or 8), both the numbers (x + m) and (x – m) are odd integers. Odd integers will not have any of these factors: 2, 4, 8, 16.

So (x + m) and (x – m) must have 1 as the only common factor. Hence their greatest common divisor must be 1.

Together, the two statements are sufficient to answer the question.

Answer (C)  This question is discussed   HERE  .

Usernamevisible · Answer

## Using Negative Remainders

Question:
Find the remainder when
1555 × 1557 × 1559
is divided by 13.
-----------------

Method 1: Normal Remainders
1555 ÷ 13 leaves remainder 8
1557 ÷ 13 leaves remainder 10
1559 ÷ 13 leaves remainder 12
So:
1555 × 1557 × 1559 ≡ 8 × 10 × 12 (mod 13)
8 × 10 × 12 = 960
960 ÷ 13 leaves remainder 11
Answer = 11
-----------

Method 2: Negative Remainders (Faster)
Instead of using large positive remainders:
8 ≡ -5 (mod 13)
10 ≡ -3 (mod 13)
12 ≡ -1 (mod 13)
So:
1555 × 1557 × 1559 ≡ (-5)(-3)(-1)
= 15(-1)
= -15
Now reduce:
-15 ≡ -2 (mod 13)
Convert to a positive remainder:
13 - 2 = 11
Answer = 11
-----------

Key Idea
If a remainder r is close to the divisor d, replace it with:
r - d
Examples with divisor 13:
12 ≡ -1
11 ≡ -2
10 ≡ -3
9 ≡ -4
8 ≡ -5
This often makes multiplication much easier.

When positive integer n is divided by 3, the remainder is 1. When n is divided by 7, the remainder is 5. What is the smallest positive integer p, such that (n + p) is a multiple of 21?

(A) 1 
 (B) 2 
 (C) 5 
 (D) 19 
 (E) 20  
 (  When positive integer n is divided by 3, the remainder is 1. When n is : Problem Solving (PS) )

Solution :  
 Given
When n is divided by 3:
n ≡ 1 (mod 3)
When n is divided by 7:
n ≡ 5 (mod 7)
Step 1: Convert to negative remainders
Remainder 1 when divided by 3 means:
n is 2 less than the next multiple of 3
So:
n ≡ -2 (mod 3)
Remainder 5 when divided by 7 means:
n is 2 less than the next multiple of 7
So:
n ≡ -2 (mod 7)
Step 2: Use the "same remainder → LCM gets same remainder" rule
Now we have:
n ≡ -2 (mod 3)
and
n ≡ -2 (mod 7)
Same remainder (-2).
Therefore:
n ≡ -2 (mod 21)
because
LCM(3,7) = 21
Step 3: Interpret
n ≡ -2 (mod 21)
means:
n is 2 less than a multiple of 21.
Examples:
19, 40, 61, 82, ...
All are 2 less than multiples of 21.
Step 4: Find p
Need:
n + p
to be a multiple of 21.
Since n is already 2 less than a multiple of 21,
add 2.
p = 2
Answer
(B) 2
Exam Brain Tag
When you see:
• remainder 1 with divisor 3
• remainder 5 with divisor 7
Notice:
1 ≡ -2 (mod 3)
5 ≡ -2 (mod 7)
Both become the same negative remainder.
Then immediately think:
Same remainder → same remainder for the LCM.
n ≡ -2 (mod 21)
Therefore add 2 to reach a multiple of 21.

Usernamevisible · Answer

## Smaller Divisor → Larger Multiple Remainders

Core Concept
Knowing the remainder with a smaller divisor does NOT uniquely determine the remainder with a larger multiple.
If:
N ≡ r (mod d)
and you are asked for the remainder when dividing by kd,
there may be multiple possible remainders.
------------------------------------------

Example 1
Given:
N leaves remainder 3 when divided by 5
N = 5k + 3
Possible numbers:
3, 8, 13, 18, 23, 28, ...
Now divide by 10:
3 → remainder 3
8 → remainder 8
13 → remainder 3
18 → remainder 8
23 → remainder 3
Possible remainders:
3, 8
----

Why Only 3 and 8?
Every number is:
(some groups of 5) + 3
When dividing by 10:
2 groups of 5 = 1 group of 10
So only the unpaired groups of 5 matter.
Possible leftovers:
0 extra groups of 5 + 3 = 3
1 extra group of 5 + 3 = 8
Cannot have 2 extra groups of 5 because they form another group of 10.
Therefore:
Possible remainders = 3, 8
--------------------------

GMAT Application
Question:
A number leaves remainder 3 when divided by 5.
What is its remainder when divided by 10?
Many students answer:
3
Wrong.
Could be:
3 (3, 13, 23, ...)
8 (8, 18, 28, ...)
Information is insufficient.
----------------------------

Example 2
Given:
N ≡ 2 (mod 4)
Possible numbers:
2, 6, 10, 14, 18, ...
Asked:
Remainder when divided by 8?
Check:
2 → remainder 2
6 → remainder 6
10 → remainder 2
14 → remainder 6
Possible remainders:
2, 6
Not unique.
-----------

General Rule
If:
N ≡ r (mod d)
and:
Asked for remainder modulo kd
Possible remainders are:
r, r + d, r + 2d, ..., r + (k − 1)d
Total possibilities = k
-----------------------

Original Example
Given:
N ≡ 1 (mod 3)
Asked:
Remainder when divided by 9
Since:
9 = 3 × 3
Possible remainders:
1
1 + 3 = 4
1 + 6 = 7
Answer:
1, 4, 7
-------

Exam Brain Tag
When you know:
Remainder with d
and are asked:
Remainder with a larger multiple of d
Do NOT look for one remainder.
Ask:
"How many possible remainders can exist?"
This is often a Data Sufficiency trap because one remainder modulo a smaller divisor can correspond to several remainders modulo a larger multiple.

Usernamevisible · Answer

## Successive Division

Core Idea
In successive division, do NOT keep dividing the original number.
After each division, divide the quotient from the previous step.
----------------------------------------------------------------

Example
Start with:
37
First divide by 4:
37 = 4(9) + 1
Quotient = 9
Remainder = 1
Now divide the quotient (9) by 5:
9 = 5(1) + 4
Quotient = 1
Remainder = 4
Final remainders:
1 and 4
-------

Structure
If:
N = d1(Q1) + r1
Then:
Q1 = d2(Q2) + r2
Then:
Q2 = d3(Q3) + r3
Always divide the previous quotient, not N.
-------------------------------------------

## Successive Division Shortcut

Question Type
Given:
Divisors
Remainders
Find the smallest number satisfying the successive divisions.
-------------------------------------------------------------

Method
Work backwards from the last remainder.
Rule:
New Number = (Current Number × Previous Divisor) + Previous Remainder
---------------------------------------------------------------------

Example
Divisors:
5, 7, 8
Remainders:
2, 3, 4
Start from the last remainder:
4
Move backward:
4 × 7 + 3 = 31
31 × 5 + 2 = 157
Smallest number = 157
---------------------

Verification
157 ÷ 5
= 31 remainder 2
31 ÷ 7
= 4 remainder 3
4 ÷ 8
= 0 remainder 4
Remainders obtained:
2, 3, 4
Correct.
--------

GMAT Exam Brain Tag
Successive division means:
Original Number → Quotient → Quotient → Quotient
Never:
Original Number → Original Number → Original Number
For "find the number" questions:
Work backwards using:
(Current Number × Divisor) + Remainder
until you reach the original number.

Question -  On dividing a certain number by 5, 7 and 8 successively, the remainder : Problem Solving (PS)

Usernamevisible · Answer

## Let's solve it exactly the way the article does: cyclicity + negative remainder.

Step 1: Find the remainder pattern of powers of 3 mod 5
Compute a few powers:
3^1 = 3 ⇒ remainder 3
3^2 = 9 ⇒ remainder 4
3^3 = 27 ⇒ remainder 2
3^4 = 81 ⇒ remainder 1
Pattern:
3, 4, 2, 1
Cycle length = 4.
-----------------

Step 2: Determine (7^11) mod 4
Because the remainder of 3^k depends only on (k mod 4).
Need:
7^11 mod 4
Use negative remainder:
7 = 8 − 1
So:
7^11 = (8 − 1)^11
When dividing by 4, every term containing 8 is divisible by 4.
Only the last term matters:
(−1)^11 = −1
Therefore:
7^11 ≡ −1 (mod 4)
which means:
7^11 ≡ 3 (mod 4)
----------------

Step 3: Go back to the cycle
For powers of 3 mod 5:
Exponent mod 4 = 1 → Remainder 3
Exponent mod 4 = 2 → Remainder 4
Exponent mod 4 = 3 → Remainder 2
Exponent mod 4 = 0 → Remainder 1
We found:
7^11 ≡ 3 (mod 4)
So the exponent is in the 3rd position of the cycle.
Therefore:
3^(7^11)
has the same remainder as:
3^3
when divided by 5.
3^3 = 27
27 ÷ 5 leaves remainder 2.
--------------------------

Answer:
2
Choice (C)
----------

Exam Brain Tag

1. Find cycle of (3^n mod 5): (3, 4, 2, 1), length 4.
2. Find (7^11 mod 4).
3. Use 7 = (8 − 1) ⇒ 7^11 ≡ (−1)^11 ≡ 3 (mod 4).
4. Exponent ≡ 3 mod 4 ⇒ use 3rd term of cycle ⇒ remainder = 2.

Interesting Number Properties

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GMAT Problem Solving (PS) Questions

Interesting Number Properties

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