Distributivity and Algebra | IT

Question 16

What could be a general rule to multiply a number by 101 and write the product in one line? Extend this rule for multiplication by 1001, 10001, ...

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Solution

We can use the distributive property of multiplication to simplify these calculations.

Step 1 — Deriving the Rule for 101

Let NN be any number we want to multiply.

We want to calculate N×101N \times 101.

We know that 101 can be written as 100 + 1.

So, N×101N \times 101 becomes N×(100+1)N \times (100 + 1).

We use the distributive property. This property lets us multiply a number by a sum.

N×(100+1)=N×100+N×1N \times (100 + 1) = N \times 100 + N \times 1

Multiplying NN by 100 means placing two zeros after NN.

For example, 3874×100=3874003874 \times 100 = 387400.

Multiplying NN by 1 means the number NN stays the same.

For example, 3874×1=38743874 \times 1 = 3874.

So, the product N×101N \times 101 is the sum of (NN with two zeros) and (NN).

Let's use the example 3874×1013874 \times 101:

3874×101=3874×(100+1)3874 \times 101 = 3874 \times (100 + 1)

=3874×100+3874×1= 3874 \times 100 + 3874 \times 1

=387400+3874= 387400 + 3874

391274\boxed{391274}

Diagram 1

Step 2 — General Rule for Multiplication by 101

From the previous step, we can form a general rule.

To multiply any number by 101:

First, write down the original number.

Next, write the same number again.

Shift this second copy two places to the right.

Add the two numbers together.

This method gives the product in one line.

Step 3 — Extending the Rule for 1001, 10001, and so on

Let us extend this idea for other numbers like 1001 or 10001.

For multiplication by 1001:

We can write 1001 as 1000 + 1.

So, N×1001=N×(1000+1)N \times 1001 = N \times (1000 + 1).

Using the distributive property, this equals N×1000+N×1N \times 1000 + N \times 1.

N×1000N \times 1000 means NN followed by three zeros.

So, we write the number NN.

Then, we write NN again, shifted three places to the right.

We add these two numbers.

For multiplication by 10001:

We can write 10001 as 10000 + 1.

So, N×10001=N×(10000+1)N \times 10001 = N \times (10000 + 1).

Using the distributive property, this equals N×10000+N×1N \times 10000 + N \times 1.

N×10000N \times 10000 means NN followed by four zeros.

So, we write the number NN.

Then, we write NN again, shifted four places to the right.

We add these two numbers.

We can observe a clear pattern here.

The number of zeros in the multiplier (like 1001 or 10001) tells us how many places to shift the second copy of the number.

For 101, there are two zeros, so we shift two places.

For 1001, there are three zeros, so we shift three places.

For 10001, there are four zeros, so we shift four places.

Answer

A general rule to multiply any number by 101 is:

• Write the number. • Write the same number again, but shifted two places to the right (like adding two zeros). • Add the two numbers. This gives the product in one line. Example: 3874×101=387400+3874=3912743874 \times 101 = 387400 + 3874 = 391274

This idea can be extended: For ×1001\times 1001:

• Write the number. • Write the same number again, shifted three places to the right. • Add them.

For ×10001\times 10001:

• Write the number. • Write the same number again, shifted four places to the right. • Add them.

So for multiplication by 101, 1001, 10001, … just write the number twice and the second copy is shifted according to the number of zeros.

More questions in IT

Q1

Context: Consider the multiplication of two numbers, say, 23 × 27.

Q. By how much does the product increase if the first number (23) is increased by 1?

Q2

Context: Consider the multiplication of two numbers, say, 23 × 27.

Q. What if the second number (27) is increased by 1?

Q3

Context: Consider the multiplication of two numbers, say, 23 × 27.

Q. How about when both numbers are increased by 1?

Q4

Context: Consider the multiplication of two numbers, say, 23 × 27.

Q. Do you see a pattern that could help generalise our observations to the product of any two numbers?

Q5

What would we get if we had expanded (a+1)(b+1)(a + 1) (b + 1) by first taking (b+1)(b + 1) as a single term? Try it!

Q6

Will the product always increase? Find 3 examples where the product decreases.

Q7

What happens when aa and bb are negative integers?

Check by substituting different values for aa and bb in each of the above cases. For example, a=5,b=8;a=4,b=5a = -5, b = 8; a = -4, b = -5; etc.

Q8

Use Identity 1 to find how the product changes when

(i) one number is decreased by 2 and the other increased by 3;

(ii) both numbers are decreased, one by 3 and the other by 4.

Q9

Verify the answers by finding the products without converting the subtractions to additions.

Q10

Expand (i) (au)(b+v)(a - u)(b + v), (ii) (au)(bv)(a - u)(b - v).

Q11

Use the following multiplications to find the product of a number with 11 in a single step.

(a) 3874×113874 \times 11 (b) 5678×115678 \times 11

Q12

Describe a general rule to multiply a number (of any number of digits) by 11 and write the product in one line.

Evaluate: (i) 94×1194 \times 11 (ii) 495×11495 \times 11 (iii) 3279×113279 \times 11 (iv) 4791256×114791256 \times 11

Q13

Can we come up with a similar rule for multiplying a number by 101?

Q14

Multiply 3874 by 101.

Q15

Use this to multiply 3874×1013874 \times 101 in one line.

Q16

What could be a general rule to multiply a number by 101 and write the product in one line? Extend this rule for multiplication by 1001, 10001, ...

Q17

Use this to find (i) 89×10189 \times 101, (ii) 949×101949 \times 101, (iii) 265831×1001265831 \times 1001, (iv) 1111×10011111 \times 1001, (v) 9734×999734 \times 99 and (vi) 23478×99923478 \times 999.

Q18

The area of a square of sidelength 60 units is 3600 sq. units (60260^2) and that of a square of sidelength 5 units is 25 sq. units (525^2). Can we use this to find the area of a square of sidelength 65 units?

Q19

What if we write 65265^2 as (30+35)2(30 + 35)^2 or (52+13)2(52 + 13)^2? Draw the figures and check the area that you get.

Q20

If aa and bb are any two integers, is (a+b)2(a + b)^2 always greater than a2+b2a^2 + b^2? If not, when is it greater?

Q21

Use Identity 1A to find the values of 1042104^2, 37237^2. (Hint: Decompose 104 and 37 into sums or differences of numbers whose squares are easy to compute.)

Q22

Use Identity 1A to write the expressions for the following.

(i) (m+3)2(m + 3)^2

(ii) (6+p)2(6 + p)^2

Q23

Expand (3j+2k)2(3j + 2k)^2 using both the identity and by applying the distributive property.

Q24

Find the general expansion of (ab)2(a - b)^2 using geometry, as we did for 55255^2.

Q25

Use the identity (ab)2(a - b)^2 to find the values of (a) 99299^2 and (b) 58258^2.

Q26

Expand the following using both Identity 1B and by applying the distributive property

(i) (b6)2(b - 6)^2 (ii) (2a+3)2(-2a + 3)^2 (iii) (7y34z)2\left(7y - \frac{3}{4z}\right)^2

Q27

Take a pair of natural numbers. Calculate the sum of their squares. Can you write twice this sum as a sum of two squares?

Try this with other pairs of numbers. Have you figured out a pattern?

Notice that 2(52+62)=(6+5)2+(65)22 (5^2 + 6^2) = (6 + 5)^2 + (6 - 5)^2.

Q28

Do the identities below help in explaining the observed pattern?

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

(ab)2=a22ab+b2(a - b)^2 = a^2 - 2ab + b^2

(a+b)2+(ab)2=(a2+2ab+b2)+(a22ab+b2)(a + b)^2 + (a - b)^2 = (a^2 + 2ab + b^2) + (a^2 - 2ab + b^2)

Adding the like terms a2+a2=2a2a^2 + a^2 = 2a^2, b2+b2=2b2b^2 + b^2 = 2b^2 and 2ab2ab=02ab - 2ab = 0, we get

2(a2+b2)=(a+b)2+(ab)22(a^2 + b^2) = (a + b)^2 + (a - b)^2

Q29

Pattern 2

9×91×1=10×88×86×6=14×27×72×2=9×510×104×4=14×6\begin{aligned} 9 \times 9 - 1 \times 1 &= 10 \times 8 \\ 8 \times 8 - 6 \times 6 &= 14 \times 2 \\ 7 \times 7 - 2 \times 2 &= 9 \times 5 \\ 10 \times 10 - 4 \times 4 &= 14 \times 6 \end{aligned}

Here is a related pattern. Try to describe the pattern using algebra to determine if the pattern always holds.

Q30

Use Identity 1C to calculate 98×10298 \times 102, and 45×5545 \times 55.

Q31

Show that (a+b)×(ab)=a2b2(a + b) \times (a - b) = a^2 - b^2 geometrically.

Q32

Context: Sridharacharya (750 CE) gave an interesting method to quickly compute the squares of numbers using Identity 1C! Consider the following modified form of this identity — a2=(a+b)(ab)+b2a^2 = (a + b)(a - b) + b^2

Q. Why is this identity true?

Q33

6.3 Mind the Mistake, Mend the Mistake

We have expanded and simplified some algebraic expressions below to their simplest forms.

(i) Check each of the simplifications and see if there is a mistake. (ii) If there is a mistake, try to explain what could have gone wrong. (iii) Then write the correct expression.

Q34

6.4 This Way or That Way, All Ways Lead to the Bay

Observe the pattern in the figure below. Draw the next figure in the sequence. How many circles does it have? How many total circles are there in Step 10? Write an expression for the number of circles in Step k.

Q35

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Q. Use this formula to find the number of circles in Step 15.

Q36

Consider the pattern made of square tiles in the picture below.

(i) How many square tiles are there in each figure? (ii) How many are there in Step 4 of the sequence? What about Step 10? (iii) Write an algebraic expression for the number of tiles in Step nn. Share your methods with the class. Can you find more than one method to arrive at the answer?

Q37

Context: Consider the pattern made of square tiles in the picture below.

Q. How many square tiles are there in each figure?

Q38

Context: Consider the pattern made of square tiles in the picture below.

Q. How many are there in Step 4 of the sequence? What about Step 10?

Q39

Context: Consider the pattern made of square tiles in the picture below.

Q. Write an algebraic expression for the number of tiles in Step nn. Share your methods with the class. Can you find more than one method to arrive at the answer?

Q40

By expanding both expressions, check that (m+n)24mn=(nm)2(m + n)^2 - 4mn = (n - m)^2.

Q41

By expanding the expressions, verify that all three expressions are equivalent. If x=8x = 8 and y=3y = 3, find the area of the shaded region.

Q42

Write an expression for the area of the dashed region in the figure below. Use more than one method to arrive at the answer. Substitute p=6p = 6, r=3.5r = 3.5, and s=9s = 9, and calculate the area.

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