✦✧✧ Question ✧✧✦ ➠ ᴅᴏɴᴛ sᴘᴀᴍ⚠️ correct answer will surely get brillianist Find the smallest number which when decreased by 11 is exactly divisible by both 420 and 168.
Answers
Answer:
To find the least number, we need to find the LCM of 36, 48, 21 and 28.
36=2×2×3×3
48=2×2×2×2×3×3
21=3×7
28=2×2×7
Thus the LCM of these numbers is 2
4
×3
2
×7=1008
The LCM of 36, 48, 21 and 28 is 1008.
When 5 is added to 1008, we get 1013.
Hence, 1013 is the least number when decreased by 5 is divisible by 36, 48, 21 and 28.
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Answer:
NOT SAME QUES
BUT SAME METHOD
Step-by-step explanation:
The given numbers are 520 and 468.
Let us find the LCM of 520 and 468
520 = 2 × 2 × 2 × 5 × 13
468 = 2 × 2 × 3 × 3 × 13
LCM of 520 and 468 = 2 × 2 × 2 × 3 × 3 ×5 × 13
LCM of 520 and 468 = 4680
The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is obtained by subtracting 17 from the LCM of 520 and 468.
The smallest number which when increased by 17 is exactly divisible by both 520 and 468 = LCM of 520 and 468 – 17
The smallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4680 – 17
The smallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4663.