**Search this site**

# 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?

**Some useful links for your career:**

- Union Public Service Commission - www.upsc.gov.in
- IIT-Kharagpur - www.iitkgp.ac.in
- Indian Statistical Institute - www.isical.ac.in
- Indian Institute of Technology Madras - www.iitm.ac.in
- Indian Institute of Management, Ahmedabad - www.iimahd.ernet.in
- Indian Institute of Mass Commission - www.iimc.nic.in
- IIT Bombay - www.iitb.ac.in
- Indian School of Mines, Dhanbad - www.ismdhanbad.ac.in
- Birla Institute of Technology, Ranchi - www.bitmesra.ac.in
- Central Institute of Fisheries Nautical and Engineering Training - www.cifnet.nic.in
- Indian Institute of Information Technology, Allahabad (Deemed University) - www.iiita.ac.in
- Central Marine Fisheries Research Institute, Kochi - www.cmfri.com
- Tata Institute of Social Sciences, Mumbai - www.tiss.edu