Friendly Numbers: What could possibly make two numbers friendly?

Well, mathematicians have decided that two numbers are considered friendly (or as often used in the more sophisticated literature, “amicable”) if the sum of the proper divisors of one equals the second and the sum of the proper divisors of the second number equals the first number.

Have a look at the smallest pair of friendly numbers: 220 and 284.

The proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110. Their sum is 1+2+4+5+10+11+20+22+44+55+110= 284.

The proper divisors of 284 are 1, 2, 4, 71, and 142, and their sum is 1 + 2 + 4 + 71 + 142 = 220.

This shows the two numbers are friendly numbers.

The second pair of friendly numbers to be discovered (by Pierre de Fermat, 1601–1665) was 17,296 and 18,416:

17,296 = 2⁴.23.47 and 18,416 = 2⁴.1,151

The sum of the proper factors of 17,296 is

1+2+4+8+16+23+46+47+92+94+184+188+368+376+752+1081+2162+4324+8648=18416

The sum of the proper factors of 18,416 is

1 + 2 + 4 + 8 + 16 + 1,151 + 2,302 + 4,604 + 9,208 = 17,296

Here are a few more friendly pairs of numbers:

1,184 and 1,210

2,620 and 2,924

5,020 and 5,564

6,232 and 6,368

10,744 and 10,856

9,363,584 and 9,437,056

111,448,537,712 and 118,853,793,424

Want to verify the above pairs’ “friendliness”?

Well, the following is one method for finding friendly numbers.

Let

a = 3.2ⁿ − 1

b = 3.2ⁿ⁻¹ − 1

c = 3².2²ⁿ⁻¹ − 1

where n is an integer greater than or equal to 2 and a, b, & c are all prime numbers. Then 2ⁿab and 2ⁿc are friendly numbers.

(Notice that for n ≤ 200, the values of n = 2, 4 and 7 give us a, b, and c to be prime.)

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Finding remainder on dividing numbers

Remainder on dividing by 3

To find the remainder of a number divided by 3, add the digits of the number and divide it by 3. So if the digits added together equal 8 then the number has a remainder of 2 since 8 divided by 3 has a remainder of 2.

For example, find the remainder on dividing 1,342,568 by 3.

The digit sum of 1342568 is 1+3+4+2+5+6+8 = 29.

Again, the digit sum of 29 = 2+9 = 11 = 2

So, the remainder will be 2.

Another example: Take 34,259,677,858.

The digit sum of the number is 3+4+2+5+9+6+7+7+8+5+8 = 64.

Now, the digit sum of 64 is 6+4 = 10 = 1.

So, the remainder is 1.

Similarly, the digit sum of 54,670,329,845 is 53 i.e. (5+3=) 8.

When we divide 8 by 3, we get remainder 2.

So the answer will be 2.

Remainder on dividing by 4

To find the remainder of a number divided by 4, take the remainder of the last 2 digits.  So if the last 2-digits are 13 then the number has a remainder of 1 since 13 divided by 4 has a remainder of 1.

Remainder on dividing by 5

To find the remainder of a number divided by 5, simply use the last digit.  If it is greater than 5, subtract 5 for the remainder.

Remainder on dividing by 7

Split the digits of the number in group of 3 starting from unit’s place. Add the alternate group and then find their difference. Divide the difference by 7 and get the remainder.

Consider the number 34568276.

Make triplets as written below starting from unit’s place
34.........568..........276
Now alternate sum = 34+276=310 and 568 and difference of these sums =568-310=258

Divide it by 7 we get remainder as 6.

Another example, consider the number 4523895099854

Triplets pairs are 4…523…895…099…854

Alternate sums are 4+895+854=1753 and 523+099=622

Difference =1131

Revise the same tripling process 1…131

So difference = 131-1=130

Divide it by 7 we get remainder as 4.

Remainder on dividing by 8

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Some Interesting Properties of Numbers

Square of Sum = Sum of Cubes

An interesting property of numbers is that, the square of sum of first ‘n’ natural numbers is equal to the sum of their cubes:

(1 + 2 + 3 +...+ n)² = 1³+ 2³+ 3³+...+ n³

Examples:

(1 + 2 + 3 + 4 + 5)²           = 225 =    1³ + 2³ + 3³ + 4³ + 5³

(1 + 2 + 3 + 4 + 5 + 6)²      = 441 =    1³+2³+3³+4³+5³+6³

(1+2+3+4+5+6+7)²             = 784 =    1³+2³+3³+4³+5³+6³+7³

Sum of all the digits of a Cube

The sum of any cube when reduced to a single digit will always be either 1, 8, or 9.

For Example:

5³ = 125 = 1+2+5 =8

9³ = 729 = 7+2+9 = 18=1+8=9

17³ = 4913=4+9+1+3=17=1+7=8

4³=64=6+4=10=1+0=1

also, they follow a cycle. The numbers 1, 8, and 9 appear in order:

1 - sum of the digits of cube of 1 would be 1

2 - would be 8

3 - would be 9

4 - would be 1 (4³ = 64 = 6+4 = 10= 1+0 = 1)

5 - would be 8 (5³ = 125 = 1+2+5 = 8)

6 - would be 9 (6³ = 216 = 2+1+6 = 9)

So on and so forth.
 

Armstrong Number

If sum of cubes of each digit of the number is equal to the number itself, then the number is called an Armstrong number.

For example,

153 = 1³ + 5³ + 3³

370 = 3³ + 7³ + 0³

371 = 3³ + 7³ + 1³

407 = 4³ + 0³ + 7³

Parasitic Number

A Parasitic number (in base 10) is a Positive natural number which can be multiplied by n by moving the rightmost digit of its decimal representation to the front.

102564 × 4 = 410256

142857 × 5 = 714285

179487 × 4 = 717948

105263157894736842 × 2 = 210526315789473684

1014492753623188405797×7=7101449275362318840579

Munchausen Number

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Multiplying any number by 11

You likely all know the 10 rule (to multiply by 10, just attach a ‘0’ after the number) but do you know the 11 rule? It is as easy! You should be able to do this one in your head for any two digit number.

Some may try this way: multiplying the number by 10, and add the original number to the result.

For example: 17 × 11 =?

17 × 10 = 170 + 17 = 187

So, 17 × 11 = 187

And, 156 X 11 =?

Add a zero to the end: 1560

Add 156 to this: 1560 + 156 = 1716

Here is a very nifty way to multiply any number by 11.

Multiplying a 2-digit number by 11

• To multiply a two-digit number by 11, just add the two digits and place the sum between the two digits.

For example, suppose you need to multiply 45 by 11. According to the rule, add 4 and 5 and place it between the 4 and 5 to get 495. It’s as simple as that.

Let’s try with 52.

• Separate the two digits in your mind (5__4).
• Notice the hole between them!
• Add the 5 and the 2 together (5+2=7)
• Put the resulting 7 in the gap 574. That's it! 11 x 54=594

The only thing tricky to remember is that if the sum of the two digits is greater than 9, then you only place the ‘units’ digit between the two digits of the number being multiplied by 11 and “carry” the ‘tens’ digit to be added to the left digit of the multiplicand.

For example 11 x 57 ... 5__7 ... 5+7=12 ... put the 2 in the hole and add the 1 from the 12 to the 5 in to get 6 for a result of 627 ... 11 x 57 = 627

Let’s try with 78 × 11. Since, 7 + 8 = 15, we place the 5 between the 7 and 8 and add the 1 to the 7, to get [7 + 1][5][8] or 858.

Multiplying a number more than 2 digit by 11

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Squaring numbers ending in 5

To square a number that ends in 5, simply

1. Multiply the ten’s (left) digit(s) by the next higher digit (itself+1)
2. Attach 25 at the end.

Example: If the number to be squared is 35;

Notice, the first digit is 3, and the next higher digit to 3 is 4.

So, the first part of the answer is 3 × 4 = 12 (first digit multiplied by next higher digit). And the last part of the answer is always 25.

So, 35² = 1225

Similarly, 75² = 5625. The answer is in two parts.

The first part is the first number; which is 7, multiplied by the next higher number that is 8; giving (7 x 8 =) 56.  The last part is always 25, giving the final answer 5625.

We can also square 3-digit numbers this way, by starting with the first two digits:

Example:      125² =?

a.  First part is 12 × 13 = 156

b.  The last part is 25.

So, 125² = 15,625

Similarly, 995² = 99 × 100 = 9900 and 25; that is 990025

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