NCERT Class 6 Solutions: Whole Numbers (Chapter 2) Exercise 2.3–Part 2

Q-4 Find using distributive property:

  1. 728×101

  2. 5437×1001

  3. 824×25

  4. 4275×125

  5. 504×35

Solution:

Equality of terms in distributive property

Understanding the Distributive Property

Equality of terms in distributive property

  1. 728×101

    =728×(100+1)

    =728×100+728×1

    =72800+728

    =73528

  2. 5437×1001

    =5437×(1000+1)

    =5437×1000+5437×1

    =5437000+5437

    =5442437

  3. 824×25

    =(800+20+4)×25

    =800×25+20×25+4×25

    =20000+500+100

    =20000+600

    =20600

  4. 4275×125

    =(4000+200+805)×125

    =4000×125+200×125+80×1255×125

    =500000+25000+10000625

    =534375

  5. 504×35

    =(500+4)×35

    =500×35+4×35

    =17500+140

    =17640

Q-5 Study the pattern:

Given the pattern
Given the pattern and find the next two steps

1×8+1=9

12×8+2=98

123×8+3=987

1234×8+4=9876

12345×8+5=98765

Write the next two steps. Can you say how the pattern works?

Solution:

The pattern works in such a way that in each step the digit added to the least significant digit is one less than previous digit. Therefore we can write the numbers as follows:

  • 123456×8+6=987648+6=987654

  • 1234567×8+7=9876536+7=9876543

Let’s understand why such a pattern emerges with an example:

  • 1234×10+4=12344 , this is true for all numbers so we can generalize it 1234n×10+n=1234nn

  • Now subtract 1234 from our example we get, or number 1234×10+41234=123441234=11110 . Applying the distributive property we get 1234×(101)+4=1234×(9)+4=11110 . In general 1234n×10+n1234n=1234n×9+n=111(ntimes)0

  • Now subtract 1234 again, we will get 1234×9+41234=1234×(91)+4=1234×(8)+4=111101234 . Now we can rewrite, 11110 – 1234 = (10000 + 1000 + 100 + 10) – (1000 + 200 + 30 + 4).

    Again using the distributive property we can write,

    (10×1000+10×100+10×10+10×1)(1×1000+2×100+3×10+4×1) ,

    which we know would be (101)×1000+(102)×100+(103)×10+(104)×1

    = 9×1000+8×100+7×10+6 = 9000 + 800 + 70 + 6=9876. Therefore, 1234×8+4=9876 generalizing this argument we can write, 1234n×8+n=90..(n1zeros)+80..(n2zeros)(10n)=9876..(10n) .

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