Page 310 - Start Up Mathematics_8 (Non CCE)

(b) 6 × B gives B as digit in ones place. This is possible if B is 2, 4, 6 or 8.

Case I: If B = 2, then Case II: If B = 4, then
A2 × 6 = 222 A4 × 6 = 444
fi (10A + 2) × 6 = 222 fi (10A + 4) × 6 = 444
fi 10A + 2 = 37 fi 10A + 4 = 74
fi 10A = 35 fi 10A = 70

35 7
fi A = = , which is 70
10 2 fi A = = 7, which is possible.
not possible as A is a digit. 10

Case III: If B = 6, then Case IV: If B = 8, then

A6 × 6 = 666 fi A8 × 6 = 888
fi (10A + 6) × 6 = 666 fi (10A + 8) × 6 = 888
fi 10A + 6 = 111 fi 10A + 8 = 148

fi 10A = 105 fi 10A = 140
105 21 fi A = 140 = 14, which is not
fi A = = , which is 10
10 2
not possible as A is a digit. possible as A is a digit.
\ The required solution is A = 7, B = 4.

Example 8: Solve MN + NM = PMP, where MN and NM are 2-digit numbers and PMP is a 3-digit number.
Solution: Here MN is a 2-digit number and NM is obtained by reversing its digits.
Now, MN + NM = PMP
fi (10M + N) + (10N + M) = PMP

fi 11M + 11N = PMP fi 11(M + N) = PMP
Now, M + N cannot exceed 18 (as per the definition of generalized form of a number, M and
N can maximum be 9 each).
So, PMP cannot exceed 11 × 18 = 198 and should be a multiple of 11.
Let’s take all the 3-digit numbers less than 198 which are multiple of 11.
110, 121, 132, 143, 154, 165, 176, 187, 198

Among these numbers only 121 is a 3-digit number whose ones and hundreds digits are
same.

\ P = 1 and M = 2
121
Now, 11(2 + N) = 121 fi 2 + N = = 11
11
fi N = 11 – 2 = 9

\ M = 2, N = 9, P = 1

Number Puzzles and Games

Example 9: Take any 3-digit number and multiply it by 13. Now multiply the product by 7 and the new
product formed by 11. What do you find? Explain the answer.

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