In the calculations shown, each letter stands for a digit.
They are used to make some two-digit numbers. The
two numbers on the left have a total of 79. What is the
total of the four numbers on the right?
AB
AD
CD + CD
79
+ CB
?
+
+AB
A) 79
B
158
C
869
D
1418
Answers
Answer:
Step-by-step explanation:
Introduction
You have studied various types of numbers such as natural numbers, whole numbers, integers and rational numbers. You have also studied a number of interesting properties about them. In previous classes, we explored finding factors and multiples and the relationships among them.
In this chapter, we will explore numbers in more detail. These ideas help in justifying tests of divisibility.
Numbers in General Form
Let us take the number 52 and write it as
52 = 50 + 2 = 10 × 5 + 2
Similarly, the number 37 can be written as
37 = 10 × 3 + 7
In general, any two digit number ab made of digits a and b can be written as
ab = 10 × a + b = 10a + b
ba = 10 × b + a = 10b + a
Let us now take number 351. This is a three digit number. It can also be written as
351 = 300 + 50 + 1 = 100 × 3 + 10 × 5 + 1 × 1
497 = 100 × 4 + 10 × 9 + 1 × 7
Similarly, In general, a 3-digit number abc made up of digits a, b and c is written as
abc = 100 × a + 10 × b + 1 × c
= 100a + 10b + c
In the same way,
cab = 100c + 10a + b
bca = 100b + 10c + a and so on.
Games with Numbers
Reversing the digits – two digit number
Minakshi asks Sundaram to think of a 2-digit number, and then to do whatever she asks him to do, to that number. Their conversation is shown in the following figure.
It so happens that Sundaram chose the number 49. So, he got the reversed number 94; then he added these two numbers and got 49 + 94 = 143. Finally he divided this number by 11 and got 143 ÷ 11 = 13, with no remainder. This is just what Minakshi had predicted.
Now, let us see if we can explain Minakshi’s “trick”.
Suppose Sundaram chooses the number ab, which is a short form for the 2-digit number 10a + b. On reversing the digits, he gets the number ba = 10b + a. When he adds the two numbers he gets:
(10a + b) + (10b + a) = 11a + 11b = 11 (a + b).
So, the sum is always a multiple of 11, just as Minakshi had claimed. Observe here that if we divide the sum by y 11, the quotient is a + b, which is exactly the sum of the digits of chosen number ab. Let us see how Sundaram explains Minakshi’s second “trick”.
Suppose he chooses the 2-digit number ab = 10a + b. After reversing the digits, he gets the number ba = 10b + a. Now Minakshi tells him to do a subtraction, the smaller number from the larger one.
If the tens digit is larger than the ones digit (that is, a > b), he does:
(10a + b) – (10b + a) = 10a + b – 10b – a = 9a – 9b = 9(a – b).
If the ones digit is larger than the tens digit (that is, b > a), he does:
(10b + a) – (10a + b) = 9(b – a).
And, of course, if a = b, he gets 0.
In each case, the resulting number is divisible by 9. So, the remainder is 0. Observe here that if we divide the resulting number (obtained by subtraction), the quotient is a – b or b – a according as a > b or a < b. You may check the same by taking any other two digit numbers.
Reversing the digits – three digit number
Think of a 3-digit number, Now make a new number by putting the digits in reverse order, and subtract the smaller number from the larger one. Divide your answer by 99.There will be no remainder!!!
In fact, Minakshi chose the 3-digit number 349. So she got:
Reversed number: 943;
Difference: 943 – 349 = 594;
Division: 594 ÷ 99 = 6, with no remainder.
Let us see how this trick works.
Let the 3-digit number chosen by Minakshi be abc = 100a + 10b + c. After reversing the order of the digits, she gets the number cba = 100c + 10b + a. On subtraction:
If a > c, then the difference between the numbers is
(100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a
= 99a – 99c = 99(a – c).
If c > a, then the difference between the numbers is
(100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a).
And, of course, if a = c, the difference is 0.
Summary
Numbers can be written in general form. Thus, a two digit number ab will be written as ab = 10a + b.
The sum of a 2-digit number and the number obtained by interchanging its digits is always divisible by 11.
The difference between a 2-digit number and the number obtained by interchanging its digits is always divisible by 9.
The general form of a 3-digit number is 100a +10b + c.
The difference between a 3-digit number and a number obtained by reversing its digits is always divisible by 99.
The general form of numbers are helpful in solving puzzles or number games.