# Solution

There are 210 laddoos, 252 imartis, and 294 balushahis.

The first and foremost thing that I would do is find the prime factors of each of these numbers.

So, 210 = 2 × 3 × 5 × 7

252 = 2 × 2 × 3 × 3 × 7

294 = 2 × 3 × 7 × 7.

So, the highest common factor = 2 × 3 × 7 = 42.

So, the highest number that divides 210, 252, and 294 = 42.

So, 210 = 42 × 5

252 = 42 × 6

294 = 42 × 7.

So, I can have 42 boxes, each containing five laddoos, such that no laddoo is left over.

Similarly, I can have 42 boxes, each containing six imartis, such that no imarti is left over.

Finally, I can have 42 boxes, each containing seven balushahis, such that no balushahi is left over.

Thus, I can have a maximum of 42 boxes, each containing five laddoos, six imartis, and seven balushahis, such that no sweets are left over.

Did I do it right?

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