Divisibility and Multiples | IT

Question 12

Context: The algebraic expressions from the previous page are:

Q. Similarly, determine and explain which of the other expressions always give even numbers. Write a couple of examples and non-examples, as appropriate, for each expression.

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Solution

We will explain why certain algebraic expressions always result in an even number, providing examples for each.

Step 1 — Sum of even products

We know that any integer multiplied by an even number is always an even number. Here, 6p6p is a product of 6 (an even number) and an integer pp, so 6p6p is always even. Similarly, 4q4q is a product of 4 (an even number) and an integer qq, so 4q4q is always even. The sum of two even numbers is always an even number.

Let us take some examples. Let p=1p = 1 and q=2q = 2. 6(1)+4(2)6(1) + 4(2) =6+8= 6 + 8 =14= 14

Let p=−3p = -3 and q=5q = 5. 6(−3)+4(5)6(-3) + 4(5) =−18+20= -18 + 20 =2= 2

Step 2 — Difference of even products

The term 10x10x is always an even number because it is 10 (an even number) multiplied by an integer xx. The term 2y2y is always an even number because it is 2 (an even number) multiplied by an integer yy. When we subtract an even number from another even number, the result is always an even number.

Let us take some examples. Let x=3x = 3 and y=4y = 4. 10(3)−2(4)10(3) - 2(4) =30−8= 30 - 8 =22= 22

Let x=−2x = -2 and y=7y = 7. 10(−2)−2(7)10(-2) - 2(7) =−20−14= -20 - 14 =−34= -34

Step 3 — Product with an even factor

An even number multiplied by any integer always gives an even result. In this expression, 8a8a is a product of 8 (an even number) and an integer aa, so 8a8a is always even. Then, we multiply this even number 8a8a by another integer bb. The final product will always be an even number.

Let us take some examples. Let a=2a = 2 and b=5b = 5. 8(2)×58(2) \times 5 =16×5= 16 \times 5 =80= 80

Let a=−1a = -1 and b=−3b = -3. 8(−1)×(−3)8(-1) \times (-3) =−8×(−3)= -8 \times (-3) =24= 24

Step 4 — Doubling an integer sum

Let the sum inside the bracket, m2+n2m^2 + n^2, be represented by an integer, say KK. Since mm and nn are integers, m2m^2 and n2n^2 are also integers. Their sum KK will therefore be an integer. The expression becomes 2K2K. Any integer multiplied by 2 always results in an even number.

Let us take some examples. Let m=3m = 3 and n=4n = 4. 2(32+42)2(3^2 + 4^2) =2(9+16)= 2(9 + 16) =2(25)= 2(25) =50= 50

Let m=−2m = -2 and n=−1n = -1. 2((−2)2+(−1)2)2((-2)^2 + (-1)^2) =2(4+1)= 2(4 + 1) =2(5)= 2(5) =10= 10

Step 5 — Multiplying by four

Let the sum inside the bracket, p+q+rp + q + r, be represented by an integer, say SS. Since p,q,rp, q, r are integers, their sum SS will also be an integer. The expression becomes 4S4S. Any integer multiplied by 4 is always an even number, because 4 itself is an even number.

Let us take some examples. Let p=1,q=1,r=1p = 1, q = 1, r = 1. 4(1+1+1)4(1 + 1 + 1) =4(3)= 4(3) =12= 12

Let p=−4,q=2,r=0p = -4, q = 2, r = 0. 4(−4+2+0)4(-4 + 2 + 0) =4(−2)= 4(-2) =−8= -8

Step 6 — Difference of even multiples

The term 12k12k is always an even number because it is 12 (an even number) multiplied by an integer kk. The term 8m8m is always an even number because it is 8 (an even number) multiplied by an integer mm. The difference between two even numbers is always an even number.

Let us take some examples. Let k=1k = 1 and m=1m = 1. 12(1)−8(1)12(1) - 8(1) =12−8= 12 - 8 =4= 4

Let k=−3k = -3 and m=2m = 2. 12(−3)−8(2)12(-3) - 8(2) =−36−16= -36 - 16 =−52= -52

Step 7 — Twice an integer sum

Let the sum inside the bracket, a+b+c+da + b + c + d, be represented by an integer, say TT. Since a,b,c,da, b, c, d are integers, their sum TT will also be an integer. The expression becomes 2T2T. Any integer multiplied by 2 always results in an even number.

Let us take some examples. Let a=2,b=5,c=−3,d=1a = 2, b = 5, c = -3, d = 1. 2(2+5+(−3)+1)2(2 + 5 + (-3) + 1) =2(7−3+1)= 2(7 - 3 + 1) =2(5)= 2(5) =10= 10

Let a=0,b=0,c=0,d=7a = 0, b = 0, c = 0, d = 7. 2(0+0+0+7)2(0 + 0 + 0 + 7) =2(7)= 2(7) =14= 14

Step 8 — Expanding and simplifying

We can simplify this expression using algebraic identities. We know that (A−B)2=A2−2AB+B2(A - B)^2 = A^2 - 2AB + B^2 and (A+B)2=A2+2AB+B2(A + B)^2 = A^2 + 2AB + B^2. Let A=xA = x and B=yB = y. So, the expression becomes (x2−2xy+y2)−(x2+2xy+y2)(x^2 - 2xy + y^2) - (x^2 + 2xy + y^2). When we remove the brackets and combine like terms, we get −4xy-4xy. Since −4-4 is an even number, the product −4xy-4xy will always be an even number, regardless of the integer values of xx and yy.

Let us simplify the expression. (x–y)2–(x+y)2(x – y)^2 – (x + y)^2 =(x2−2xy+y2)−(x2+2xy+y2)= (x^2 - 2xy + y^2) - (x^2 + 2xy + y^2) =x2−2xy+y2−x2−2xy−y2= x^2 - 2xy + y^2 - x^2 - 2xy - y^2

−4xy\boxed{-4xy}

Let us take some examples. Let x=3x = 3 and y=1y = 1. −4(3)(1)-4(3)(1) =−12= -12

Let x=−2x = -2 and y=5y = 5. −4(−2)(5)-4(-2)(5) =40= 40

Answer

(i) 6p+4q6p + 4q (ii) 10x–2y10x – 2y (iii) 8a×b8a \times b (iv) 2(m2+n2)2(m^2 + n^2) (v) 4(p+q+r)4(p + q + r) (vi) 12k–8m12k – 8m (vii) 2(a+b+c+d)2(a + b + c + d) (viii) (x–y)2–(x+y)2(x – y)^2 – (x + y)^2

More questions in IT

Q1

Evaluate each expression and write the result next to it. Do you notice anything interesting?

Q2

Now, take four other consecutive numbers. Place the '+' and '-' signs as you have done before. Find out the results of each expression. What do you observe?

Q3

Repeat this for one more set of 4 consecutive numbers. Share your findings.

Q4

Do these patterns occur no matter which 4 consecutive numbers are chosen? Is there a way to find out through reasoning?

Hint: Use algebra and describe the 8 expressions in a general form.

Q5

Now take any 4 numbers, place '+' and '-' signs in the eight different ways, and evaluate the resulting expression. What do you observe about their parities?

Repeat this with other sets of 4 numbers.

Q6

Is there a way to explain why this happens?

Hint: Think of the rules for parity of the sum or difference of two numbers.

Q7

Context: Now, let us see what happens when a negative sign is switched to a positive sign.

Q. Replace any negative sign in the expression a+b−c−da + b - c - d with a positive sign and find the difference between the two numbers.

Q8

Context: Replace any negative sign in the expression a+b−c−da + b - c - d with a positive sign and find the difference between the two numbers.

Q. What do you conclude from this observation?

Q9

Is the phenomenon of all the expressions having the same parity limited to taking 4 numbers? What do you think?

Q10

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We know how to identify even numbers. Without computing them, find out which of the following arithmetic expressions are even.

Q11

Using our understanding of how parity behaves under different operations, identify which of the following algebraic expressions give an even number for any integer values for the letter-numbers.

Q12

Context: The algebraic expressions from the previous page are:

Q. Similarly, determine and explain which of the other expressions always give even numbers. Write a couple of examples and non-examples, as appropriate, for each expression.

Q13

Write a few algebraic expressions which always give an even number.

Q14

Pairs to Make Fours

Take a pair of even numbers. Add them. Is the sum divisible by 4?

Try this with different pairs of even numbers. When is the sum a multiple of 4, and when is it not? Is there a general rule or a pattern?

Q15

When will two even numbers add up to give a multiple of 4?

This problem is similar to the question of identifying when adding two numbers will result in an even number. Can you see this?

There are three cases to examine:

Q16

Look at the following expressions and the visualisation. Write the corresponding explanation and examples.

Q17

Always, Sometimes, or Never

We examine different statements about factors and multiples and determine whether a statement is 'Always True', 'Sometimes True', or 'Never True'.

We know that the sum of any two multiples of 2 is also a multiple of 2.

  1. If 8 exactly divides two numbers separately, it must exactly divide their sum.

Statement 1 is always true. Determine if it is true with subtraction.

Q18

Examine each of the following statements, and determine whether it is 'Always true', 'Sometimes true', 'Never true'.

  1. If a number is divisible by both 9 and 4, it must be divisible by 36.

  2. If a number is divisible by both 6 and 4, it must be divisible by 24.

Q20

Context: Let us consider another expression, 5k−25k - 2, and see the values it takes for different values of kk.

Numbers that leave a remainder of 33 when divided by 55 can also be seen as 22 less than multiples of 55; 5k−25k - 2, where k≥1k \ge 1.

Q. Are there other expressions that generate numbers that are 33 more than a multiple of 55?

Q21

Similarly, explain using algebra why the divisibility shortcuts for 5, 2, 4, and 8 work.

Q22

Look at each of the following statements. Which are correct and why?

(i) If a number is divisible by 9, then the sum of its digits is divisible by 9.

(ii) If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.

(iii) If a number is not divisible by 9, then the sum of its digits is not divisible by 9.

(iv) If the sum of the digits of a number is not divisible by 9, then the number is not divisible by 9.

Q23

The shortcut to find the divisibility by 3 is similar to the method for 9. A number is divisible by 3 if the sum of its digits is divisible by 3. Explore the remainders when powers of 10 are divided by 3. Explain why this method works.

Q24

Using these observations, can you tell whether the number 462 is divisible by 11?

Q25

Context: This alternating pattern of one more than 11 and one less than 11 continues for higher place values. Since 400 contains 4 hundreds, 400 is 4 more than a multiple of 11 (396+4396 + 4). Since 60 contains 6 tens, 60 is 6 less than a multiple of 11 (66−666 - 6). Since 2 contains 2 units, 2 is 2 more than a multiple of 11, i.e., 2=(0+2)2 = (0 + 2). Using these observations, can you tell whether the number 462 is divisible by 11?

Q. What could be a general method or shortcut to check divisibility by 11?

Q26

If this difference is 11 or a multiple of 11, what does that say about the remainder obtained when the number is divisible by 11?

Q27

Using this shortcut, find out whether the following numbers are divisible by 11. Further, find the remainder if the number is not divisible by 11.

(i) 158 (ii) 841 (iii) 481 (iv) 5529 (v) 90904 (vi) 857076

Q28

Is this method similar to or different from the method we saw just before?

Q29

Fill in the following table. Find a quick way to do this?

Q30

More on Divisibility Shortcuts

Divisibility Shortcuts for Other Numbers

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Q31

Will checking its divisibility by its factors 2 and 3 work? Use the shortcuts for 2 and 3 on these numbers and divide each number by 6 to verify— 38, 225, 186, 64.

Q32

How about checking divisibility by 24? Will checking the divisibility by its factors, 4 and 6, work? Why or why not?

Q33

What property do you think this digital root will have? Recall that we did this while finding the divisibility shortcut for 9.

Q34

Between the numbers 600 and 700, which numbers have the digital root: (i) 5, (ii) 7, (iii) 3?

Q35

Write the digital roots of any 12 consecutive numbers. What do you observe?

Q36

Now, find the digital roots of some consecutive multiples of (i) 3, (ii) 4, and (iii) 6.

Q37

What are the digital roots of numbers that are 1 more than a multiple of 6? What do you notice?

Try to explain the patterns noticed.

Q38

I’m made of digits, each tiniest and odd, No shared ground with root #1—how odd!

My digits count, their sum, my root— All point to one bold number’s pursuit— The largest odd single-digit I proudly claim.

What’s my number? What’s my name?

Q39

Solve the cryptarithms given below:

Q40

Try this now: GH × H = 9K.

This means a 2-digit number multiplied by a 1-digit number gives another 2-digit number in the 90s. Observe the letters corresponding to the units digits in this cryptarithm. Pick the solution to this question from the options given below: 11 × 9 = 99, 12 × 8 = 96, 46 × 2 = 92, 24 × 4 = 96, 47 × 2 = 94, 31 × 3 = 93, 16 × 6 = 96.

Q41

Solve the following:

(i) UT × 3 = PUT

(ii) AB × 5 = BC

(iii) L2N × 2 = 2NP

(iv) XY × 4 = ZX

(v) PP × QQ = PRP

(vi) JK × 6 = KKK

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