23. Find the lowest natural number which when dividend by 112, 140, and 168 leaves a remainder 8 in each case. (a) 1688 (b) 1699 (c) 1234 (d) 14560.
Answers
Answer:
The correct answer is 1688
Step-by-step explanation:
Let us find the lowest number which when divided by 112, 140, and 168 separately, leaves no remainder. Then add 8 to it.
So, 112 = 2×2×2×2×7.
140 = 2×2×5×7.
168 = 2×2×2×3×7.
Out of these, let us list out the maximum number of times 2, 3, 5, and 7 (i.e., all multiplicants) used in multiplication.
That is, 2 is used 4 times (in obtaining 112).
3 is used 1 time (in obtaining 168).
5 is used 1 time (in obtaining 140).
7 is used 1 time (in obtaining 112, 140 and 168).
Now, let us perform (2×4)×(3×1)×(5×1)×(7×1).
That is, 8×3×5×7 = 840.
Now, let us divide 840 by 112, 140, and 168 individually.
840/112 = 7.5
840/140 = 6
840/168 = 5.
Since 840 is not an integral multiple of 112, (but of 140 and 168), our intention is to get 15 in the place of 7.5. (And, while doing so, integral status of divisions by 140 and 168 would not be altered even though answers get doubled).
Let us perform (840 × 2) to get the integer answer of (7.5 × 2 = ) 15, when divided by 112.
That is, 840 × 2 = 1680 is the lowest number which is the lowest common multiplier of 112, 140, and 168.
In order to get the remainder 8 when divided by 112, 140, and 168, we need to add 8 to 1680.
Therefore, 1688 is the answer.