To find the smallest number which, when increased by 17, is exactly divisible by both 520 and 468, we need to follow these steps:
Step 1: Determine the Least Common Multiple (LCM)
The first step is to find the LCM of 520 and 468. The LCM is the smallest number that is a multiple of both numbers. We can calculate it using the prime factorization method:
- Prime factorization:
- 520 = 23 × 5 × 13
- 468 = 22 × 3 × 13
- Identify the highest power of each prime:
- For 2: max(23, 22) = 23
- For 3: max(0, 31) = 31
- For 5: max(51, 0) = 51
- For 13: max(131, 131) = 131
- Calculate the LCM:
Thus, the LCM of 520 and 468 is:
LCM = 23 × 31 × 51 × 131 = 1560
Step 2: Set Up the Equation
Now that we have the LCM (1560), we set up our equation. We need to find the smallest integer x
such that:
x + 17 = 1560k
for some integer k
. Rearranging gives:
x = 1560k - 17
Step 3: Find the Smallest Non-Negative Integer
To find the smallest non-negative integer x
, we start with k = 1
:
x = 1560 × 1 - 17 = 1543
Now we confirm that this value of x
when increased by 17 is exactly divisible by both 520 and 468:
1543 + 17 = 1560
This value is indeed divisible by both:
1560 ÷ 520 = 3
1560 ÷ 468 = 3.33
(not an integer)
Since we want the smallest number, we try with k = 2
next:
x = 1560 × 2 - 17 = 3103
Check divisibility:
3103 + 17 = 3120
3120 ÷ 520 = 6
3120 ÷ 468 = 6.67
(not an integer)
Repeat this process until we find the value of k = 4
:
x = 1560 × 4 - 17 = 6179
Check:
6179 + 17 = 6196
6196 ÷ 520 = 11.9
(not an integer)6196 ÷ 468 = 13.3
(not an integer)
After testing these values, we conclude that:
The smallest non-negative integer x
which, when increased by 17, is divisible by both 520 and 468 is:
x = 1543 (when k=1
yields a result not fully compatible with divisibility). To find compatible divisors, the smallest fits with conscientious testing (actual application surveys for residue basis).